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THEORY  AND  CALCULATION 


OF 


CANTILEVER  BRIDGES. 


BY  R.  M.  WILCOX,  PH.  B., 

Instructor  in  Civil  Engineering  in  Lehigh  University. 


NEW  YOEK 

>.  VAN  NOSTRAND  COMPANY 

$3  MURBAY  AND   27   WARREN    STREET 

1898 


Copyright,  1898, 

BY 
D.  VAN  NOSTBAND  COMPANY. 


PREFACE. 


THIS  volume  replaces  the  original  No.  25 
of  Van  Nostrand's  Science  Series,  bear- 
ing the  title  "  Theory  and  Calculation  of 
Continuous  Bridges,"  by  Prof.  Mansfield 
Merriman,  which  was  published  in  1875. 

The  continuous  girder,  though  exten- 
sively built  in  Europe  prior  to  1875,  has 
now  gone  entirely  out  of  use,  except  for 
revolving  draw-bridges,  and  the  cantilever 
bridge  has  taken  its  place.  Indeed,  the 
modern  cantilever  bridge  is  simply  a 
continuous  girder  with  the  chords  cut, 
a  form  of  construction  which  lacks  most 
of  the  theoretical  objections  of  its  ances- 
tor, and  at  the  same  time  possesses  the 
very  great  advantage  over  simple  trusses 
of  erection  without  false  work. 

This  book  has  been  written  with  the 
object  of  presenting  as  clearly  as  possible 
the  theory  and  methods  of  calculating  the 
stresses  in  the  trusses  of  cantilever  bridg- 


IV.  PREFACE. 

es.  Both  highway  and  railroad  structures 
are  discussed.  In  each  case  a  sufficient 
number  of  the  stresses  have  been  worked 
out  to  illustrate  the  application  of  the 
methods,  and  the  stresses  in  all  the  mem- 
bers  are  given  in  tables. 


B.M    WILCOX 


SOUTH  BETHLEHEM,  PA., 
March,  1898. 


CONTENTS. 


CHAPTER  I.— Introduction. 

AKT.  1 .     History  of  Cantilever  Bridges. 
"     2.     Classification. 


CHAPTER  II.— Highway  Bridges. 

AET.  3.     Definitions. 

1 '    4.     Dead  Load. 

* l    5.     Reactions  Due  to  Dead  Load. 

"     6.     Shear  and  Shear  Diagrams. 

'  *     7.     Moment  and  Moment  Diagrams. 

"  8.  Cantilevers  with  Horizontal  Chords; 
Stress  in  Web  Members. 

"  9.  Cantilevers  with  Horizontal  Chords; 
Stress  in  Chord  Members. 

* '  10.     Cantilevers  with  One  Chord  Inclined. 

"11.  Shears  and  Moments  Due  to  Concen- 
trated Live  Load. 

"  12.  Max.  +  and  —  Shear  Due  to  Uniform 
Live  Load. 

"13.  Max.  +  and  —  Moment  Due  to  Uniform 
Live  Load. 

"  14.  Cantilever  with  Horizontal  Chords,  Uni- 
form Live  Load  Stresses. 


71.  CONTENTS. 

ABT.  15.  Snow  Load  and  Snow  Load  Stresses* 

"     16.  Stresses  Due  to  Wind. 

* '     17.  False  Members  for  Purposes  of  Erection. 

'  *     18.  Final  Max.  and  Min.  Stresses. 

CHAPTEE  III.— Railroad  Bridges. 

ABT.  19.  Loads  on  Cantilever  E.  E.  Bridges. 

"    20.  Eeaction  Due  to  Dead  Load. 

"     21.  Stresses  Due  to  Dead  Load. 

"    22.  Live  Load. 

"    23.  Live  Load  Stresses. 

"    24.  Wind  Load  Stresses. 


CHAPTER     I. 


INTRODUCTION. 


ARTICLE  1. — HISTORY. 

THE  cantilever  bridge  is  a  develop- 
ment of  the  continuous  girder ;  in  fact  it 
is  the  continuous  girder  with  the  chords 
cut  and  hinged  (properly)  at  the  points  of 
reversion  of  flexure. 

The  first  real  practical  type  of  canti- 
lever bridge  consisted  of  two  sets  of  logs 
projecting  out  from  the  two  opposite  shores 
of  a  stream  and  the  space  between  the 
ends  of  these  arms  spanned  by  other  logs 
or  beams.  Such  a  bridge  was  built  in 
Thibet  about  240  years  ago.  For  a  de- 
scription see  R.  R.  Gazette,  1882,  p.  2. 

A  book  entitled  "A  Treatise  on  Bridge 
Architecture/7  by  Thomas  Pope,  published 
in  New  York  in  1811,  sets  forth  a  scheme 
for  bridging  the  Hudson  river.  It  was 
called  "  Pope's  Flying  Pendant  Lever 


2 


Bridge/'  and  contained  the  principles  of 
the  cantilever  bridge ;  but  Pope's  ideas 
were  decidedly  erroneous  as  regards  the 
stresses. 

At  a  time  when  tubular  bridges  and 
continuous  girders  were  in  favor, — about 
1850, — it  was  suggested,  by  Edwin  Clark, 
that  the  chords  in  continuous  girders  be 
severed  at  the  points  of  contrary  flexure, 
and  the  central  portion  be  hung  at  those 
points.  This  plan  though  not  carried  into 
practical  operation  until  some  twenty-five 
years  later,  was  nevertheless  the  essential 
principle  of  the  modern  cantilever  bridge. 
In  1833  M.  A.  Canfield  built  a  bridge  at 
Paterson,  N.  J.,  which  is  claimed  to  be 
the  first  cantilever  bridge  ever  built  in 
America.  In  1876-77  C.  Shailer  Smith 
built  the  "  Kentucky  River  Bridge,"  300 
feet  above  water.  A  suspension  bridge 
was  originally  intended,  and  towers  were 
built  for  that  purpose.  The  bridge  was 
built  out  from  the  shore  panel  by  panel 
until  the  towers  were  reached,  and  then 
continued  on  until  connections  were  made 
at  the  middle.  Then,  in  order  to  avoid 


alternate  stresses  which  would  be  pro- 
duced if  the  bridge  was  perfectly  stiff, 
the  chords  were  cut  on  the  shore  arms 
near  the  piers.  This  bridge  is  located  on 
the  Kentucky  river,  about  112  miles  from 
Cincinnati.  For  full  description  see 
Transactions  of  the  Am.  Soc.  C.  E.,  Nov. 
1878. 

In  1867-68  Prof.  W.  P.  Trowbridge, 
of  the  School  of  Mines  of  Columbia  College, 
New  York,  conceived  of  and  executed  a 
plan  for  the  first  long-span  cantilever  in 
America.  It  was  designed  to  span  the 
East  river  opposite  76th  street,  New  York, 
and  involved  the  construction  of  two  im- 
mense masonry  piers  135  feet  high  placed 
on  BlackwelPs  Island.  On  top  of  these 
masonry  piers  it  was  intended  to  place 
iron  towers  150  feet  higher.  For  full 
description  of  the  proposed  bridge  see 
Eng.  News,  Dec.  29,  1883. 

"  The  Niagara  Cantilever  Bridge  "  was 
begun  in  April,  and  completed  in  Decem- 
ber, 1883.  It  was  considered  a  wonderful 
piece  of  engineering,  both  in  the  rapidity 
of  construction  and  obstacles  overcome. 


It  was  designed  by  C.  C.  Schneider,  M. 
Am.  Soc.  C.  E.,  and  built  by  the  "  Cen- 
tral Bridge  Works"  of  Buffalo,  N.  Y. 
The  principal  dimensions  are :  Length  over 
all  910  feet,  each  cantilever  375  feet,  and 
central  span  120  feet.  It  has  two  points 
of  support  25  feet  apart  at  the  piers,  which 
are  simply  iron  towers.  The  structure 
carried  two  lines  of  railroad  299  feet  above 
the  surface  of  the  water  of  Niagara  river. 
A  paper  on  this  bridge  is  to  be  found  in 
vol.  XIV  of  the  Transactions  of  Am.  Soc., 
C.  E. 

The  next  cantilever  bridge  of  impor- 
tance built  in  America  was  the  "  St.  John 
River  Bridge."  It  was  opened  for  traffic 
in  September,  1885,  and  was  another  ex- 
ample of  rapid  construction.  It  had  the 
following  general  dimensions :  Total  length 
812J  feet,  two  cantilevers  of  287  and  382 
feet  respectively,  and  a  central  span  of 
143J  feet.  A  full  account  of  this  bridge 
is  to  be  found  in  R.  R.  Gazette,  1885, 
p.  691. 

"  The  Louisville  Bridge,"  over  the 
Ohio  river,  connecting  the  cities  of  Lou- 


isville,  Ky.,  and  New  Albany,  Ind.,  con- 
sists of  two  cantilever  spans  480  and  483 
feet  long  respectively,  separated  by  a  con- 
tinuous span  of  360  feet ;  two  anchor  spans 
of  260  feet  each,  a  swing  span  of  370  feet, 
and  a  fixed  span  on  the  New  Albany  side 
of  240  feet — making  a  total  length  of  2453 
feet.  The  distance  from  the  under  side  of 
the  trusses  to  the  water  is  95  feet.  It 
was  built  in  1886,  by  the  Union  Bridge 
Co.  under  trying  difficulties,  and  was 
made  of  open-hearth  steel.  See  Eng.  News, 
Nov.  27,  1886. 

"  The  Poughkeepsie  Bridge/7  over  the 
Hudson  river  at  Poughkeepsie,  N.  Y.,  has 
a  total  length  (not  including  viaduct  ap- 
proaches) of  3093  feet.  It  is  212  feet  above 
high  water  and  consists  of  five  spans 
of  continuous  and  cantilever  trusses.  It 
was  built  by  the  Union  Bridge  Co.,  in 
1887-88.  The  foundations  for  the  piers  of 
this  bridge  were  very  deep,  one  being  129 
feet  below  the  surface  of  the  river.  See 
R.  R.  Gazette,  July  1,  1887,  and  also  Eng. 
News,  Oct.  29,  1887. 

The  "  Philadelphia  Cantilever  Bridge," 


6 

over  the  Schuylkill  river  at  Market  street, 
completed  in  1888,  is  about  409  feet  long 
and  77  feet  wide,  and  consists  of  two  canti- 
lever spans  166  feet  lOf  indies,  and  one 
central  span  76  feet  long. 

The  "  Great  Forth  Bridge "  was  com- 
menced in  1881  and  completed  in  1890. 
li  crosses  the  "Firth  of  Forth"  in  Scotland, 
and  consists  of  three  gigantic  cantilevers 
connected  by  two  central  spans  each  350 
feet  long.  The  middle  cantilever  is  1620 
feet  long,  and  rests  on  two  supports  260 
feet  apart.  The  other  two  cantilevers  are 
each  1505  feet  long,  and  rest  on  two  sup- 
ports of  145  feet  apart.  The  total  length 
of  the  bridge  is,  therefore,  5330  feet.  This 
length  does  not  include  the  approaches, 
which  in  themselves  are  immense  struc- 
tures. The  maximum  distance  between 
piers  is  1700  feet,  the  longest  span  in  the 
world.  The  clearance  of  the  central  spans 
above  high  water  is  150  feet.  A  full  his- 
tory and  description  of  this  bridge  is  given 
in  London  Engineering,  of  1890,  p.  213. 

Other  important  cantilever  bridges 
have  been  built,  principal  among  which  in 


America  may  be  mentioned  the  "  Bed  Rock 
Cantilever  Bridge  "  in  California,  and  the 
"Memphis  Bridge"  at  Memphis,  Term. 
The  former  was  built  in  1890,  and  its  main 
span  is  660  feet  long.  See  R.  R.  Gazette, 
April  25,  1890,  and  Eng.  News,  Sept.  27 
and  Oct.  4,  1890. 

The  "  Memphis  Bridge "  was  opened 
for  traffic  in  1892.  Largest  span  790  feet. 
See  Eng.  News,  May  12,  1892. 

ARTICLE  2.— CLASSIFICATION. 

A  cantilever  bridge,  as  defined  in  the 
Century  dictionary,  consists  of  bracket- 
shaped  beam  trusses  extending  inward 
from  their  supports  and  connected  at  the 
middle  of  the  span,  either  directly  or  by 
an  intermediate  span  of  ordinary  construc- 
tion. 

This  arrangement  is  shown  in  Fig.  I, 
and  is  the  simplest  type  of  the  modern 
cantilever  bridge.  Various  modifications 
of  the  arrangement  of  the  trusses  exist  in 
cantilever  bridges,  but  all  contain  the 
principle  of  the  bracket  or  arm  supporting 


a  weight^  which  is  kept  in  equilibrium  by 
a  counter- weight  or  reaction. 

Cantilever  bridges  are  arbitrarily  divid- 
ed into  two  general  classes,  depending 
upon  the  arrangement  of  the  supports. 
The  first  includes  those  which  have  two 
points  of  support  at  the  pier,  as  is  shown 
in  Fig.  xxm.  The  " Niagara  Cantilever'7 
and  "  Great  Forth  "  bridges  are  examples 
of  this  class.  The  second  includes  those 
cantilever  bridges  which  at  the  pier  are 
supported  at  a  single  point.  Fig.  vil 
represents  the  arrangement  of  the  reac- 
tions for  the  second  class,  and  the  "  St. 
John  River  "  and  "  Louisville  "  bridges  are 
good  examples  of  it.  The  calculations  of 
the  reactions  for  the  two  cases  is  quite 
different,  as  will  be  seen  by  reference  to 
the  formulas  in  Articles  21  and  5  respect- 
ively. 


CHAPTEE  II. 


HIGHWAY    BRIDGES. 


ARTICLE  3.  —  DEFINITIONS. 

The  following  definitions  of  shore  arm, 
river  arm,  and  central  span  apply  gener- 
ally to  both  classes  of  cantilevers,  but  par- 
ticularly to  the  truss  arrangement  repre- 
sented by  the  Niagara  cantilever  bridger 
in  which  there  are  two  piers  and  two 
abutments. 

Fig.  vil.  shows  this  arrangement  of 
trusses,  and  reference  to  it  will  make  the 
definitions  clearer. 

Shore  arm  is  that  part  of  the  bridge  in- 
cluded between  the  abutment  and  pier,  or 
AF. 

River  arm  is  that  part  included  between 
the  pier  and  central  span,  or  FJ. 

Central  span  is  a  simple  truss  supported 
by  the  ends  of  the  river  arms.  Only  one 


10 


half  of  the  central  span  is  shown  in  Fig. 
VII. 

All  forces  acting  upward  are  to  be  taken 
as  positive,  and  all  forces  acting  downward 
negative.  Thus  a  positive  reaction  is  one 
acting  upward,  while  a  negative  reaction 
is  downward. 

Maximum  stress  means  the  greatest 
possible  stress,  either  positive  or  negative, 
that  can  occur  in  a  member.  Minimum 
stress  means  the  least  possible  stress  of 
the  same  nature  as  the  maximum  stress, 
or  if  possible  the  greatest  stress  of  the 
opposite  kind.  Maximum  and  minimum 
represent,  therefore,  the  greatest  range  of 
stress. 

Shear  diagrams  are  drawn  to  represent 
the  distribution  of  shear  throughout  the 
bridge  due  to  the  position  of  the  load 
shown,  while  moment  diagrams  represent 
the  distribution  of  moments  for  the  par- 
ticular position  of  the  load  shown. 

The  plus  sign  placed  before  a  stress 
means  tension,  and  the  minus  sign  com- 
pression. 


11 
ARTICLE  4. — DEAD  LOAD. 

The  problem  of  deducing  a  general  for- 
mula for  dead  load  in  a  cantilever  bridge, 
in  order  to  calculate  the  dead  load  stresses, 
is  one  very  difficult  to  solve,  either  theo- 
retically or  empirically.  There  are  so  many 
different  forms  of  cantilevers,  varying  in 
so  many  ways,  that  each  one  seems  to  be 
a  distinct  problem  in  itself.  With  such 
conditions  to  contend  with,  it  seems  al- 
most impossible  to  derive  a  formula  for 
dead  load. 

No  satisfactory  formula  for  dead  load  in 
cantilever  bridges  has  ever  been  found,  to 
the  author's  knowledge,  until  very  re- 
cently. In  a  little  book  called  "  De 
Pontibus,"  by  J.  A.  L.  Waddell  (N.  Y., 
John  Wiley  &  Sons,  1898)  is  presented 
a  formula  or  diagram  for  dead  load.  From 
this  the  dead  load  for  each  apex  of  shore 
and  river  arms  can  be  found  by  means  of 
what  is  called  a  "  percentage  curve."  This 
curve  is  plotted  from  values  taken  from 
a  number  of  typical  cantilever  bridges, 
and  represents  the  ratio  of  the  dead  apex 


12 

load  of  any  panel  of  the  shore  and  river 
arms,  and  the  dead  apex  load  of  the  sus- 
pended or  central  span.  It  checks  with 
remarkable  precision  the  estimated  weight 
of  the  proposed  North  River  Bridge  at 
New  York.  Although  Mr.  Waddell  does 
not  guarantee  it  to  be  accurate  for  all 
forms  of  cantilevers,  nevertheless,  suitable 
modifications  of  it  can  probably  be  so 
made,  as  it  seems  to  be  based  on  the  right 
principle. 

In  the  absence  of  any  formula,  the  only 
way  to  get  the  dead  load  is  to  weigh  the 
material,  or  get  the  actual  shop  weights. 
This  is  laborious,  and  involves  the  calcu- 
lation of  the  stresses  in  certain  members  : 
say  the  end  panel  of  the  river  arm,  due  to 
half  the  dead  weight  of  the  central  span, 
live  load  on  central  span,  the  effect  of 
wind,  together  with  an  assumed  weight 
of  the  members  themselves.  If  the  bridge 
is  a  highway  bridge,  a  stress  due  to  snow 
load  should  be  included.  With  this  maxi- 
mum stress  the  members  considered  are 
designed  (rather  roughly  at  first),  and 
their  weight  compared  with  the  assumed 


13 

dead  weight.  If  there  is  but  slight  differ- 
ence between  the  assumed  and  actual 
weights  of  the  members,  all  well  and  good ; 
but  if  too  great  a  difference  exists  between 
them,  the  work  should  be  repeated  to  the 
extent  necessary  for  close  agreement  in 
assumed  and  actual  weights. 

This  book  is  not  intended  to  explain  the 
method  of  designing  bridges,  but  to  show 
how  to  calculate  the  stresses  in  the  mem- 
bers of  a  cantilever  bridge :  hence,  it  is  of 
little  importance  whether  the  results  ob- 
tained are  the  stresses  caused  by  the  act- 
ual weight  of  the  bridge  and  the  possible 
weights  which  may  act  upon  it.  The  dead 
apex  loads,  therefore,  will  be  assumed, 
and  such  values  taken  as  to  give  simple 
numerical  computation. 

ARTICLE  5. — REACTIONS  DUE  TO 
DEAD  LOAD. 

Let  w  =  load  per  linear  foot. 
R!  =  shore  reaction. 
R8  =  river  reaction. 
/    =  length  of  shore  arm* 


14 


m  =  length  of  river  arm. 
n  =  length  of  central  span. 
W  =  total  weight  on  bridge. 


j*—/-^ 

1  i 

i 

<r-m—  ^k  —  w  —  ^ 

1              i 

^  R  1  ^ 

H 

^\            "        " 

.  I 

Since  one  half  the  weight  of  the  central 
span,  w  n,  is  supported  at  the  end  of  the 
river  area,  the  reaction  R2  can  be  found  by 
taking  B:  as  the  origin  of  moments,  then 

R2  *  +  ~  (I  +  w)  +  *  n  (?  +  w)2  =  -0,  or 


2 


wn(l 

~ 


n  } 


But  Bt  +  R2  =  w    /  +  m  +  W 


and  Bi  =  w(l  +  m  +  g)  —  R-i  .....  (2) 

By  taking  the  end  of  the  river  arm  as 
the  origin  of  moments,  the  moment  of  the 
forces  on  the  left  is 

)*  —  Ei  (Z  -h  m)  =  0...(3) 


15 

and  R,   =    >tg^  +  MI)'-'B«m........(4) 

I  -f-  »w 

These  equations  are  sufficient  to  deter- 
mine the  reactions  in  any  cantilever  with 
supports  arranged  as  shown  in  Fig.  I. 

RI  may  have  a  positive,  negative,  or  zero 
value,  depending  upon  the  relative  length 
of  I,  m  and  n. 

The  criterion  that  RI  shall  equal  zero  is 
expressed  by  the  equation  72  -f-  m2  +  m  n 
-  0,  from  which  I  -  V  m1  +  m  n. 
When  I  is  greater  than  V  m'2  -j-  m  n,  the 
value  of  Rj  is  greater  than  zero,  or  positive^ 
and  acts  upward ;  also,  when  I  is  less  than 
V  m2  +  m  n,  the  value  of  Ri  is  negative, 
and  acts  downward. 

ARTICLE  6.— SHEAR  DUE  TO  DEAD  LOAD. 

Since  the  shear  in  any  section  of  a 
beam  is  equal  to  the  algebraic  sum  of  the 
vertical  forces  on  the  left  of  that  section, 
it  follows  that  the  shear  in  the  shore  arm 
at  any  section  distant  x  from  Ri  (see  Fig.  l) 
may  be  expressed  by  the  equation  RI  — 


16 

w  x.  For  the  river  arm  the  expression 
for  shear  in  any  section  distant  x  from 
Ri  is,  RI  +  R8  —  w  x.  The  shear  in  any 
section  of  central  span  distant  x  from  its 

left   end   is    ^-^  —  w  x. 

£ 

The  distribution  of  shears  due  to  dead 
load  for  the  different  sections  through- 
out, for  the  case  when  I  is  greater  than 
y  m2  -f-  m  n  is  represented  by  the  dia- 
gram of  Fig.  n. ?  R]  being  positive. 


.  ii 


Fig.  ill  shows  the  distribution  of  shears 
for  the  case  when  I  is  less  than  V  m2  4-  m  n^ 
or  when  Rt  is  negative. 


17 


-----  1  ----  *h  —  TO  —   ----  n  ---  - 


? 

i 

i 

V 

i 

\L/"              \. 

r 

TE' 

1 

lilt 

1 

i 
| 

IflHirmL^ 

Fig.  Ill 

ARTICLE  7. — DEAD  LOAD  BENDING 
MOMENT. 

In  order  to  find  the  moment  at  any  sec- 
tion of  the  shore  and  river  arms,  it  is  neces- 
sary first  to  find  the  values  of  B2  and  Rt 
from  equations  (1)  and  (2).  The  moment 
at  any  section  of  the  shore  arm  distant  x 
from  Ri  is  represented  bv  the  equation 

w  x2 
M  =  Ri  x 2~-  ' 

The  point  of  maximum  moment  is  where 
the  vertical  shear  equals  zero.  Put  the 
equation  for  shear  Rt  —  wx  equal  to  zero, 


18 

solve  for  #,  and  substituting  this  value  of 
x  in  the  above  equation  will  give  the  maxi- 
mum moment  in  shore  arm. 

When  RI  is  positive  there  exists,  in  the 
shore  arm,  an  inflection  point  or  point  at 
which  the  moment  changes  from  positive 
to  negative,  and  is  sometimes  called  the 
point  of  reverse  flexure.  To  find  where 

w  x2 
this  point  is,   put  R:  x ^—     equal   to 

zero,  and  solve  for  x. 

The  moment  in  the  river  arm  at  any 
section  distant  x  from  Ra  is  expressed  by 


This  moment  is  always  negative.  The 
same  result  should  be  obtained  if  the  mo- 
ment of  the  forces  on  the  right  of  section 
is  taken,  or 

,..        w  n  x        w  x*      ,  .     ,,      ,. 

JM  =  — ^ h  — ~— ,  where  x  is  the  dis- 
tance from  section  to  the  end  of 
river  arm.  This  is  often  a  simpler  equa- 
tion to  use  in  computation  than  the 
former,  in  which  RI  may  be  positive 
or  negative  according  as  I  is  greater 


19 


or  less  than  A/m2  +  in  n  .  This  should 
be  determined  and  the  proper  sign  given 
to  Rj  in  the  equation  of  moment  of  forces 
on  left.  In  the  central  span  the  moment 
is  found  just  as  in  the  case  of  a  simple 
beam.  The%  equation  of  moments  at 
section  distant  x  from  its  lift  end  is, 
..  r  w  n  x  w  x* 

Mz    ~2 2~" 

The  moment  diagram  for  the  case 
when  I  is  greater  than  vm2  -f-  m  n,  or  when 
R,  is  positive,  is  shown  in  Fig.  IV. 


K- *- 


I                         1 

1 

V                 •    s 

\/              SJ 

f*              JR< 

i                 ! 

1                                 1 

i  ^•rrrniiinillllirnTT*^ 

1 

1    ^rrr""^  '  '    ,  ;"  ^^^t^ 

When  R:  is  negative,  Fig.  V  represents 
the  distribution  of  moments. 


20 


. 1 


1 

n 

i 

1 

/                   \             \ 

X              \! 

f 

Ri                      |    Rz 

i 

| 

1 

1 

1 

• 

. 

I 

| 

^rmTTmTTTT^rrr^' 

^^^i     IIIIIIF^ 

Fig.  "V 


ARTICLE  8.— CANTILEVERS    WITH  HORI- 
ZONTAL CHORDS;   STRESSES  IN  WEB 
MEMBERS  DUE  TO  DEAD  LOAD. 

The  rule  for  finding  the  stress  in  the 
web  members  of  a  truss  with  horizontal 
upper  and  and  lower  chords  is  as  follows  : 
Pass  a  section  cutting  the  member,  the  stress 
in  which  is  to  b&  found,  and  multiply  tlw  shear 
in  the  section  l>y  the  secant  of  the  angle  which 
the  member  makes  ivith  the  vertical. 

Let  the  cantilever  shown  in  Fig.  VI 
have  length  of  shore  arm  equal  to  100  feet, 
river  arm  80  f eet,  central  span  80  feet  and 


21 

depth  16  feet.  Then  sec#  =  1.18.  Let 
uniform  dead  load  equal  250  pounds  per 
linear  foot,  or  5000  pounds  per  panel. 


•p          w  m  (I  +  m)  +  w  (I  -f  w)f 
R'  ~~  -21- 

R   ._  250  X  80  x  180  +  250  x  180' 
8  200 

=  4-58  500  pounds. 

+  m  + 

R,  =  250  (100  +  80  +  40)  —  58  500 

E,  =  —  3500  pounds. 

This  shows  that  RI  is  a  negative  reac- 
tion ',  that  is,  it  acts  downward. 

The  stress  in  a  A  equals  (R,  —  P0) 
sec.  B,  or  a  A  =  +  (3500  +  2500)  1.18  = 
+  7080  pounds. 

A  b  =  —  (3500  +  2500)  1.18  =  —  7080. 

B  c  =  —  (3500  +  2500  +  5000)  1.18 
=  —  12  980  pounds. 


22 

F<7  =  (—  R,  +  R2  —  P0  —  P,  —  etc., 
.........  P8)  sec:0.  ( 

F#  =  (—  3500  +  58  500  —  2500  —  5 
X  5000)  1.18  =  +  32  450  pounds. 

Ij  -.=  (—  3500  +  58  500  —  2500  —  8  x 
5000)  1.18  =  +14  750  pounds. 

•3j  =  *^sec0-=  7500  x  1.18  =  —  8850 

pounds. 

Enough  of  the  members  have  been  taken 
to  show  how  the  stresses  are  calculated 
for  web  members  throughout  the  bridge 
due  to  dead  load. 


ARTICLE   9.  —  CANTILEVERS  WITH  HORI- 

ZONTAL CHORDS  STRESSES  IN 

CHORD  MEMBERS. 

There  are  two  methods  of  finding  the 
stress  in  the  chord  members  of  trusses 
with  horizontal  chords  :  first,  by  the 
"  Method  of  Moments  "  ;  and,  second,  by 
the  method  of  "  Chord  Increments." 

The  method  of  chord  increments  does 
not  hold  good  when  one  of  the  chords  is 
inclined  j  so  the  method  of  moments  will 


23 

be  used  in  illustrating  the  calculation  of 
chord  stress. 

Draw  a  section  cutting  three  members,  take 
the  origin  of  moments  at  the  intersection  of  the 
two  members  cut,  other  than  the  one  in  which 
the  stress  is  to  be  found ;  then  state  the  equa- 
tion of  moments  between  flie  stress  and  tlie  ap- 
plied forces  on  the  lift  of  the  section,  and  solve 
for  the  unknown  stress. 

It  is  necessary  at  first  to  calculate  the 
reactions  R8  and  RI,  just  as  was  done  in 
Art.  8.  If  same  example,  data,  etc.,  be 
taken  in  this  case  as  used  in  the  last  ar- 
ticle, then  R2  =  58  500  and  Rt  =  —  3500 
pounds. 

Let  it  be  required  to  find  the  stress  in 
be,  see  Fig.  VI.  Pass  a  section  cutting 
A  B,  B  b  and  b  c,  and  take  the  origin  of 
moments  at  B. 

The  equation  of  moment  is  then 
(_Ri_Po)30  —  P1  x  10  +  bcx  16  =  0 
-  6000  X  30  —  5000  Xl0  +  6cxl6  =  0 
and  b  c  =  —  14  375  pounds. 

Take  the  member  D  E.  The  origin  of 
moments  is  at  e  and  the  equation  is 


6000  x  80  —  15  000  X  40  +  D  E  x  16  =  0 
and  D  E  =  +  51  875  pounds. 

The  origin  of  moments  for  the  chord 
member  fg  is  at  F,  and  the  equation  of 
moments  of  the  forces  on  the  left  and  the 
stress  in/#  is 
(—  R!  —  2500)  110  +  R2  x  10  —  25  000 

X  50  +fg  x  16  =  0 
—  6000  x  110  +  58  500  x  10  —  25  000 

X  50  +fg  X  16  =  0 
and/#  =  —  82  813  pounds. 

The  moment  equation  may  express  the 
moment  of  the  forces  on  the  right  instead 
of  those  on  the  left  and  it  is  often  a  sav- 
ing of  labor  to  express  it  in  that  way ;  as, 
for  example,  to  find  the  stress  in  H  I  the 
origin  is  at  i,  and  the  equation  of  moments 
is  H I  x  16  =  12  500  x  20 
H I  =  250  000  -f-  16  =  +  15  625  pounds. 

Stressing  =  125QQX1Q  =  +  7313  pounds. 
16 

The  examples  given  are  sufficient  to 
show  how  the  stress  in  any  member  of  a 
highway  cantilever  with  horizontal  chords, 
due  to  dead  load,  can  be  found. 


25 

ARTICLE   10. — CANTILEVERS   WITH  ONE 
CHORD  INCLINED.    STRESS  IN  MEM- 
BERS DUE  TO  DEAD  LOAD. 

^ 

A  favorite  form  of  cantilever  bridge  is 
that  in  which  one  of  the  chords  is  inclined. 
When  this  arrangement  of  the  chords 
exists,  the  principle  that  the  stress  in  any 
web  member  is  equal  to  the  shear  in  the 
sections,  multiplied  by  the  secant  of  the 
angle  which  the  member  makes  with  the 
vertical  no  longer  holds  true,  for  the  reason 
that  the  inclined  chord  member  takes  up 
a  part  of  the  shear. 

Let  cantilever  shown  in  Fig.  VII  have 
length  of  shore  arm  equal  to  100  feet, 


E       F,       G       H 


river  arm  80  feet,  central  span  80  feet, 
and  distance  apart  of  trusses  16  feet ;  B  b 
=  20  feet,  F  /  =  24  feet,  I  i  =  21  feet 
and  K  Jc  =  21  feet.  Let  the  dead  load 


per  linear  foot  be,  500  pounds,  air  on  the 
upper  chord. 

Since  the  length  of  the  arms  is  the  same 
and  the  load  per  linear  foot  double  that  of 
the  examples  given  in  Art.  8,  Ra  =•  + 
58  500  x  2  =  +  117  000  and  R,  =  —  3500 
X  2  =  —  7000  pounds. 

Taking  b  as  the  center  of  moments,  the 
stress  in  A  B  is  given  by  the  equation 
A  B  X  20  —  Ri  x  20  —  5000  x  20  =  0 
and  A  B  =  B  C  =  +  12  000  pounds. 

B  &  =  —  10  000  pounds,  or  the  weight  of 
the  apex  load  that  comes  upon  it. 

For  the  stress  in  A  5,  take  the  center  of 
moments  at  B.  Then  A  b  X  14.142  = 
12  000  x  20  or  A  b  =  —  16  960  pounds. 

For  the  stress  in  the  member  &  C,  pass 
a  section  through,  cutting  B  C,  &  C  and  b  c ; 
take  the  origin  of  moments  at  the  inter- 
section of  B  C  and  It  c,  which  is  420  feet 
to  the  left  of  C  and  the  lever  arm  of  &  C 
is  296.98  feet.  The  equation  of  moments 
of  the  forces  on  the  left  of  the  section  is 
6C  X  296.98  =  12  000  x  380  +  10  000  x  400 
and  b  C  — '+  28  740  pounds. 

The  stress  in  be  is  found  by  taking  the 


origin  of  moments  at  C.  The  lever  arm 
of  fee  is  20.97  feet  giving  the  equation  fee  X 
20.97  — 12  000  x  40  —  10  000  x  20  =  0 
and  I  c  =  —  32  400  pounds. 

The  origin  of  moments  for  C  c  is  at  the 
intersection  of  C  D  and  b  c,  or  420  feet  to 
the  left  of  C,  and  the  equation  gives 
— Cc  X  420  + 12  000  X  380  +  20  000  X  410 
and  C  c  =  —  30  380  pounds. 

A  sufficient  number  of  the  members 
have  been  taken  to  illustrate  the  method 
of  calculating  the  stress  due  to  uniform 
load  in  the  members  of  the  shore  and  river 
arms. 

The  member  F/  may,  however,  offer 
some  difficulty  if  treated  according  to  the 
method  shown.  If  the  section  be  passed, 
cutting  efj  fg  and  F/,  the  solution  becomes 
very  simple  by  placing  the  vertical  com- 
ponent of  the  stress  in  ef,  fg  and  F/J  equal 
to  the  reaction  R2.  This  equation  will 
contain  only  one  unknown  quantity  F/? 
since  ef  and  fg  can  be  computed  by  the 
method  of  moments,  and  R2  is  known. 

The  equation  is : 


28 

orF/=R8-(Ve/+V/», 
in  which  Ve/and  V/^r  represent  the  vei*ti- 
cal  components  of  the  stress  in  ef  and  fg 
respectively.  These  stresses,  taken  from 
the  table,  are  133  600  and  133  500  pounds, 
and  vertical  components  of  them  6680  and 
6675  pounds.  The  above  equation  for  F/ 
reduces  then  to 

F/=  117  000  —  (6680  +  6675)  = 
— 103  600  pounds. 

The  stress  in  the  members  of  the  cen- 
tral span  are  calculated  just  like  those 
of  a  simple  deck  truss.  Since  it  has  its 
chords  parallel,  the  stress  in  any  web 
member  is  equal  to  the  shear  multiplied 
by  the  secant  of  8. 

The  stress  in  all  the  members  of  the 
cantilever  truss  shown  in  Fig.  VII,  due  to 
dead  load,  are  calculated  in  the  manner 
shown,  and  the  stresses  given  in  the  fol- 
lowing table.  The  object  in  giving  the 
tables  of  stresses  complete,  throughout 
the  book,  is  to  have  them  serve  as  an- 
swers to  any  self-selected  problem  that  the 
student  may  take.  For  instance,  if  the  stu- 
dent wishes  to  test  his  ability  in  working 


29 

out  the  stress  in  any  member  not  already 
given  he  may  take  for  example  F  Gr 
and  apply  the  same  principles  as  used  in 
finding  the  stress  in  A  B,  and  verify  his 
result  from  the  table. 

If  the  student  will  pursue  this  course 
it  will  be  found  to  be  of  very  great  help 
to  him  in  better  understanding  the  prob- 
lems. 


Dead  Load  Stresses  for  Cantilever  shown 
in  Fig.  VII. 


Member. 

Stress 
in  Pounds. 

Member. 

Stress 
in  Pounds. 

AB=:BC 

+  12000 

kl 

+  14280 

CD 

+  32380 

Eb 

-  10000 

DE 

+  60000 

bC 

+  28820 

EF 

+  94000 

Cc 

-  30400 

FG 

+  91300 

cD 

+  40120 

GH 

+  54550 

Dd 

-  39000 

HI  =  IJ 

+  23800 

dE 

+  50450 

JK 

-  14280 

Ee 

-  47300 

KL 

—  19050 

eF 

+  6000 

Ab 

-  16960 

F/ 

-103650 

be 

-  32460 

F# 

+  64090 

cd 

—  60000 

gG 

-  50440 

de 

-  94040 

Gh 

+  54700 

ef 

-133600 

m 

-  42280 

fg 

-133500 

Hi 

+  44650 

gh 

-  91400 

il 

1000 

hi 

-  54600 

m 

-  15000 

iJ 

-  34530 

Kl 

+  6900 

JJc 

+  20700 

U 

-  10000 

31 

ARTICLE    11.  —  SHEARS   AND    MOMENTS 
DUE  TO  CONCENTRATED  LIVE  LOAD. 

In  addition  to  the  dead  and  snow  load, 
cantilever  bridges  are  subjected  to  a  live 
load  stress.  This  live  load  consists,  in  the 
case  of  highway  bridges,  of  foot  people, 
horses  and  wagons,  electric  cars,  etc. 

In  order  to  determine  the  maximum 
and  minimum  stress  in  all  the  members 
affected  by  this  load,  involves  a  knowledge 
of  the  proper  position  of  live  load  to  pro- 
duce it.  What,  then,  is  the  possible  ar- 
rangement of  the  live  load  ?  Unlike  the 
dead  load,  the  live  load  may  occupy  a 
part  of,  or  different  parts  of,  the  bridge  at 
the  same  time ;  it  may  also,  like  the  dead 
load,  cover  the  entire  bridge  at  once. 

It  is  necessary,  therefore,  to  consider  all 
possible  arrangements  of  the  live  load, 
and  to  find  that  position  for  it  which  will 
produce  the  maximum  and  minimum  shear 
and  moment  for  any  section. 

Consider  first,  one  concentrated  load,  P, 
which  is  in  the  nature  of  an  electric  car  or 
heavily  loaded  wagon. 


32 

(a)     For  a  load  P  on  the  shore  arm, 


(5) 


The  effect  is  just  like  a  load  P  on  a 
simple  beam,  and  the  distribution  of  shears 
and  moments  is  as  shown  in  Fig.  Vill. 


Fig. 


(6)     For  a  load  P  on  river  arm, 
R  ,  =  -P-',  and  R,= 

I 


(6) 


and  the  shears  and  the  moments  are  dis- 
tributed as  shown  in  Fig.  ix. 


33 


L£ 


rz 


---  n  ---  - 


~\ 


TH    I 


.  IX 


(c)     For  load  P  on  central  span, 


1=_P.      ,  and  K,=  I- 

n       I  n\    I 


Fig.  X 


34 


and  Fig.  x.  shows  distribution  of  shears 
and  moments. 

An  examination  of  the  shear  diagrams 
in  the  three  cases  (a),  (b)  and  (c)  shows 
that  the  maximum  positive  shear  in  any 
section,  due  to  a  load  P,  occurs  when  the 
load  is  placed  just  to  the  right  of  the  sec- 
tion, and  that  the  maximum  negative  shear 
occurs  when  the  load  is  placed  just  to  the 
left  of  the  section. 

An  examination  of  the  moment  diagrams 
shows  that,  for  case  (a),  the  moment  is 
positive,  and  is  a  maximum  for  any  sec- 
tion in  lj  when  P  is  over  the  section ;  that, 
for  case  (&),  a  negative  moment  is  pro- 
duced in  the  shore  arm  and  negative  in 
river  arm,  both  increasing  as  m'  increases, 
and  having  a  maximum  value  when  m1 
=  m,  or  when  P  is  placed  at  the  end 
of  the  river  arm  5  and  that,  in  case  (c), 
a  negative  moment  is  produced  in  both 
the  shore  and  river  arms,  and  a  posi- 
tive moment  in  the  central  span  j  also 
that  the  shore  and  river  arm  moments  are 
a  maximum  when  P  is  placed  at  the  left 
end  of  the  central  span,  and  a  maximum 


35 


in  any  section  of  the  central  span  when 
P  is  over  the  section. 

The  above  conclusions  are  given,  in  con- 
densed form,  in  the  following  table : 

Table  showing  Position  of  Load  P  to  give  Maxi- 
mum Positive  and  Negative  Moments. 


Case  (a). 
P  ou  Shore 
Arm. 

Case  (6). 
P  on  River 
Arm. 

Case  (c). 
P  on  Central 
Span. 

Moment  in  I. 

f      Max. 
,  j  when  P  is 
~n    over  the 
t   Section. 

f      Max. 
—  •{  when  P  is 
^  at  m'=m 

f      Max. 
—  4  when  P  is 
^  at  n'=n. 

Moment  in  m. 

0 

f      Max. 
—  •{  when  P  is 
[at  m'=m 

(      Max. 
—  \  when  P  is 
(  at  n'  =n. 

Moment  in  n. 

0 

0 

f      Max. 
,    (  when  P  is 

"*"  |       over 
L   Section. 

Table  showing  Position  of  Load  P  for  Maximum 
Positive  and  Negative  Shear*. 


Max.  -f  Shear. 


Max.  —  Shear. 


Shore  Arm. 

P  just  to  right  of 
Section. 

P  just  to 
Section. 

left  of 

River  Arm. 

P  at  end  of  River 

None. 

Arm. 

Central  Span. 

P  just  to  right  of 
Section. 

P  just  to 
Section. 

left  of 

36 


ARTICLE  12. — MAXIMUM  -f-  AND  —  SHEAR 
DUE  TO  UNIFORM  LIVE  LOAD. 

The  diagrams  of  shears  and  moments  in 
the  preceding  article  represent  the  effect 
of  a  single  load  P. 

The  effect  of  any  number  of  loads,  as 
P0  P2,  Pa,  etc.,  may  be  shown  in  the  same 
manner,  the  resulting  diagram  being  the 
same  as  the  combined  diagrams  for  each 
load  taken  separately, 

When  these  loads  act  sufficiently  close 
together  and  are  of  the  same  intensity,  the 
result  is  a  uniformly  distributed  load. 
Such  a  load  is  represented  in  highway 
bridges  by  a  mass  of  foot-people  moving 
in  a  continuous  or  broken  line  across  the 
bridge.  It  is  necessary,  therefore,  to  con- 
sider the  effect  of  such  load,  and  the  posi- 
tion or  possible  arrangement  of  it  to  give 
the  maximum  +  and  —  shear  and  moment 
at  any  section. 

The  live  load  in  highway  bridges  may 
consist  of  both  concentrated  and  uniform 
load,  acting  at  the  same  time  on  different 
parts  of  the  bridge.  This  combination  is 


37 


not,  however,  generally  made  in  highway 
bridges,  as  the  result  is  not  as  injurious  as 
that  due  to  the  full  uniform  load.  If  such 
a  combination  should  be  desired,  the  rules 
of  Art.  11,  12  and  13  are  to  be  followed. 

The  shearing  effect  of  uniform  live  load 
will  now  be  considered. 

For  any  section  in  the  shore  arm,  the 
greatest  positive  shear  occurs  when  the 
load  is  so  placed  as  to  give  to  RI  the 
greatest  possible  positive  value,  with  no 
load  on  the  left  of  the  section  to  subtract 
from  it. 

Any  load  on  the  river  arm  and  central 
span  causes  a  negative  reaction  at  RI; 
therefore,  the  maximum  positive  shear  in 
any  section  of  the  shore  arm  occurs  when 
the  shore  arm  is  covered  with  the  uniform 
load  to  the  right  of  the  section,  which 
makes 


(8) 


and  gives  a  shear  diagram,  as  shown  in 
Fig.  XL 


38 

i- 


In  the  river  arm,  tne  maximum  positive 
shear  for  any  section  occurs  when  the  load 
covers  the  central  span  and  the  river  arm 
to  right  of  the  section. 

m1 
ivm1 


I 


and  R2==%±D+_ 

L*l 


Since  V— — R!+R2?  substituting  for  RI 
and  R2  their  values  as  given  above,  and 
reducing,  gives 

Ywn  i        /  /m 

=       -\-wrn' (9) 

which  equation  is  also  the  algebraic  sum 
of  the  vertical  forces  on  the  right  of  the 
.section. 


39 

The  position  of  load  for  maximum  posi- 
tive shear,  for  any  section  of  river  arm,  is 
shown  in  Fig.  XII. 


In  order  to  obtain  the  greatest  negative 
shear  in  any  section  of  the  shore  arm,  it  is 
apparent,  from  the  equation  V  =  R] — wx, 
that  Rj  should  have  the  greatest  negative 
value  possible.  Since  RI  is  negative  due 
to  the  live  load  on  the  river  arm  and  cen- 
tral span,  it  is  evident  that  the  proper 
position  of  live  load  to  fulfill  the  condition 
is,  to  cover  the  river  arm  and  central  span 
with  the  uniform  load,  and  also  the  shore 
arm  to  the  left  of  the  section. 

The  value  of  RI  is  found  from  the  equa- 
tion 


40 


wmn 


I  21  ~       21 

and  the  shear  from 

V  =  Ri— wx (10) 

The  position  of  load  and  shear  diagram 
is  shown  in  Fig.  xm. 


Kig.  XIII 

In  the  river  arm  there  is  no  negative 
shear  due  to  live  load.  In  other  words, 
there  is  no  possible  arrangement  of  the 
load  to  produce  a  negative  shear  in  the 
river  arm.  A  load  on  left  of  section  gives 


The  maximum  positive  shear  in  central 
span  for  any  section,  occurs  when  load  is 


41 


on  the  right  of  the  section;  maximum 
negative,  when  central  span  is  loaded  on 
the  left  of  section,  just  as  in  the  case  of  a 
simple  beam. 

Table  giving  Positions  of  Uniform  Live  Load  to 
Give  Max.  Positive  and  Negative  Shears. 


Max.  +  Shear. 

Max.  —  Shear. 

Shore  Ann. 

Load  to  cover  Shore 
Arm  right  of  Section. 

Load  on  River  Arm 
and  Central  Span,  also 
Shore  Arm,  left  of 
Section. 

River  Arm. 

Load  on  Central 
Span  and  River  Arm, 
right  of  Section. 

None. 

Central  Span. 

Right  of  Section. 

Left  of  Section. 

ARTICLE  13. — MAXIMUM    POSITIVE    AND 

NEGATIVE  MOMENT  DUE  TO 

UNIFORM  LIVE  LOAD. 

The  position  of  'live  load  to  produce 
maximum  moment  in  any  section  of  the 
shore  arm  is  when  the  live  load  covers  the 
entire  shore  arm. 

The   equation   of  bending   moment  at 


any  section  distant  x  from  the  left  end  is 
RI  x  — 


wx* 


wl  • 


Substituting  for  Rj  its  value  —  in  the 

u 

above  equation  gives 

nx/r          WlX  WX* 


If  any  live  load  is  placed  on  the  river 
arm  or  central  span,  its  effect  is  to  pro- 
duce a  negative  value  in  RI,  thereby  de- 
creasing the  value  of  M. 

The  diagram  in  Fig.  XIV  shows  the  dis- 
tribution of  maximum  positive  moments 
for  the  shore  arm. 


K  —  OG- 

H 

v%%2% 

'ww/twtfMww/ 

^R 

i 

! 

"R2 

I 

! 

| 

i 

nHIHIU!rrtTrr«. 

i 

! 

J 

r,        TTT'V 

43 

There  is  no  position  of  the  live  load 
that  will  give  a  positive  moment  in  the 
river  arm ;  the  moment  in  the  river  arm 
is  always  negative. 

The   central  span  being  like  a  simple 
beam,  its  maximum  positive  moment  for 
any  section  occurs  when  it  is  fully  loaded, 
as  shown  in  Fig.  XV,  in  which 
wnx       ivx2 
'2 


-««-    WHJU  'IV  JU 


.(12) 


u*  1  > 

*~  1 

i           k%%% 

%%%%%%% 

a, 

- 

1  R 

LfTTTnfll 

"TZhTT^ 

Fig.  XV 


The  maximum  negative  moment  in  the 
shore  arm  will  evidently  occur  when  the 
load  is  so  placed  as  to  give  the  greatest 
negative  value  for  Ri,  and  to  have  no  load 


44 


on  the  shore  arm  to  cause  a  positive  value 
for  Ri.  To  produce  this  result,  the  river 
arm  and  central  span  should  be  fully  loaded 
and  no  load  on  the  shore  arm. 

The  diagram  of  moments  is  shown  in 
Fig.  xvi,  and  the  value  of  the  moment  is 
found  from 

M  -  —  -Ria (13) 


H--O?—  »j                                                  ! 

^  11  — 

1             I                              Wftffiw/'/-                                   '• 

I"1  i       r 

i   ! 

i 
i 

i 
l 

The  position  of  load  to  produce  maxi- 
mum negative  moments  in  river  arm  is  of 
course  that  which  will  produce  the  great- 
est negative  value  of  RI.  This  will  occur 
when  the  river  arm  and  central  span  are 
loaded,  with  no  load  on  shore  arm,  as  in 
Fig.  XVI.  The  equation  of  moments  for 


45 

any  section  in  river  arm  distant  x  from 
is  M  =  —  Rj  (I  -f-  x)  — ^-^-  +  R2  x. 


The  moment  at  the  section  due  to  that 
part  of  the  load  on  the  left  of  the  section 

is  zero  ;  consequently,  the  quantity  —  -— 

-i 

in  the  above  equation  becomes  zero  if  the 
load  is  placed  on  the  central  span  and  on 
the  right  of  the  section  only  in  the  river 
arm.  The  effect  is  the  same,  and  the 
above  equation  is  simplified  to 

M  =  —  R1  (I  +  jc)  +  Rt  x (14) 

and  the  diagram  of  moments  for  this  posi- 
tion is  shown  in  Fig.  XVII. 


There  is  no  negative  moment  possible 
in  the  central  span,  since  it  is  a  simple 
truss. 

The  following  table  gives,  in  condensed 
form,  the  position  of  live  load  for  max. 
positive  and  negative  moments  in  the  dif- 
ferent parts  of  cantilevers. 

Table  Showing  Position  of  Uniform  Live  Load  to 
Give  Maximum  +  and  —  Moments. 


Max.  -\-  Moment. 

Max.  —  moment. 

Shore  Arm  (0- 

Load  on  entire  Shore 
Arm. 

Load  on  entire 
River  Arm  and  Cen- 
tral Span. 

Kiver  Arm  (m). 

No  -f-  moment 
possible. 

Load  on  Central 
Span  and  River  Arm, 
right  of  Section. 

Central  Span  (ri) 

Load  on  entire  Cen- 
tral Span. 

No  —  moment 
possible. 

ARTICLE  14. — CANTILEVER  BRIDGE  WITH 

HORIZONTAL  CHORD;  STRESSES  DUE 

TO  UNIFORM  LIVE  LOAD. 

To  illustrate  method  of  calculating 
stresses  in  a  cantilever  due  to  live  load, 
let  same  example  as  given  in  Article  3 


be  taken.  Shore  arm  100  feet,  river  arm 
80  feet,  central  span  80  feet,  and  depth  of 
truss  16  feet.  Sec#  =  1.18.  Let  live 
load  be  taken  at  70  pounds  per  square 
foot  of  floor  surface.  Assuming  the 
bridge  to  be  30  feet  wide,  1)he  weight 
per  linear  foot  for  one  truss  is  70  X  15  = 
1050  pounds.  This  multiplied  by  the 
panel  length  20  feet  gives  21  000,  or  say 
20  000  pounds,  as  the  live  panel  load  which 
is  all  applied  to  the  chord  that  supports 
the  floor  system. 


H i     j     K  ! 

VvAAi 


To  find  the  stress  in  A  a  caused  by  the 
maximum  positive  and  maximum  negative 
shear.  For  maximum  positive  shear,  load 
covers  shore  arm  on  right  of  section,  (see 
table  in  Art.  12).  R,  =  2  X  20  000  = 
40  000  pounds. 

Stress  in  Aa  =  40  000  X  1.18  =  —  47  200 
pounds.  For  maximum  negative  shear, 
oad  covers  river  arm,  central  span,  and 


48 

shore  arm  left  of  section. 

K  _  60  000  x  40+ 50  000  x  80 

~100~ 

Rl=r— 64  000  pounds. 
Shear  =  V  =  —  Rt  =  —  64  000  pounds. 
Stress  in  Aa  =  64  000  X  1.18=  +  75  520 
pounds. 

For  stress  in  Qh  due  to  maximum  posi- 
tive shear,  the  live  load  covers  central 
span  and  river  arm  on  the  right  of  section. 
The  panel  points  loaded  are  h,  i,  j,  k  and  I. 

Th  n  -R  — 4Q QQQ X 5Q+5Q QQQ x 8Q 
100 

R4  =  —  60  000. 
From  (2)  Ri  +  Rf=W, 

therefore  —  60  000  +  R2  =  90  000  and  R2  = 
+ 150  000  pounds. 

V  in  Grh  =  —60  000+150  000=90  000 
and  stress  inG/&=90  000  x  1.18  =  +106  200 
pounds. 

There  is  no  negative  shear  possible  in 
river  arm  5  theref ore,  no  compressive  stress 
mOfc 

For  the  stresses  in  C  D  the  maximum 
positive  moment  will  be  first  considered, 


49 


aud  this  occurs  by  reference  to  table  in 
Art.  13,  when  live  load  covers  the  entire 
shore  arm.  R,  =  2  x  20  000  =  40  000 
pounds,  and  center  of  moments  is  at  d, 
then 

n-n  _  40  000  X  60  —  40000x30 

IT" 

-  _  75  000  pounds. 
Maximum  negative  moment  for  section 
through  C  D  occurs  when  river  arm  and 
central  span  is  loaded. 
For  this  position  of  load, 

T?     _  60  000  x  40  +  50  000  X  80 

T00~ 
and  Ri  =  —  64  000 

CD=6*«»X60=+340000         ^ 
16 

Take  the  lower  chord  member  of  the 
river  arm,  ij.  There  is  no  positive  mo- 
ment possible,  so  the  maximum  negative 
moment  alone  will  be  considered.  This 
takes  place  when  live  load  covers  central 
span  and  river  arm  on  right  of  section. 

T.      50  000  V  80 

Ki  =  — -~p —  :=  —  40  000  pounds. 

R2  rr  -f  90  000  pounds. 


50 


With  center  of  moments  at  J,  stress  in  ij 
equals 

-  40  OOP  X(1Q°  +  7Q)  +  90  000X70 

16 
ij  ~  —  31  250  pounds. 

The  following  is  perhaps  a  shorter  and 
more  rapid  method  of  calculating  the  reac- 
tions in  a  cantilever  truss,  due  to  live  Ioad7 
than  the  one  just  used. 

Let  P0,  P],  P2,  P8,  etc.,  be  the  panel 
loads  at  the  apex  points,  a,  &,  c,  d,  etc.,  of 
the  truss  shown  in  Fig.  XVIII.  Then,  since 
5/5  of  P0is  supported  by  R17  4/5  of  P}  goes 
to  Rj  and  8/5  of  P2  etc.,  goes  to  the  same 
reaction,  a  table  of  coefficients  can  be 
formed  giving  the  part  of  each  load  sup- 
ported by  Rj  and  R2. 

If,  then,  the  load  is  placed  in  the  proper 
position  for  maximum,  positive,  and  nega- 
tive shear  or  moments,  the  reaction  R1? 
due  to  the.  loads  on  the  required  panel 
points  is  found  by  adding  together  the 
product  obtained  by  multiplying  each 
panel  load  by  its  coefficient  in  the  column 
Rr  In  the  same  way  the  value  of  R2  can 
be  found.  This  applies  to  loads  on  the 


51 


river  arm  and  central  span,  as  well  as  to 
loads  on  the  shore  arm. 


Load. 

Part  Supported 
by  KL 

Part  Supported 
by  Ha. 

PO 

+  1-      XPo 

o.  XP. 

Pi 

+  0.8  X  PI 

+  0.2  X  Pi 

P8 

4-  0.6  etc. 

+  0.4  etc. 

PS 

+  0.4 

+  0.6 

p, 

+  0.2 

+  0.8 

^5 

0. 

+  1. 

Pe 

-0.2 

+  1.2 

P, 

-0.4 

+  1.4 

PS 

-0.6 

+  1.6 

P9 

—  0.8 

+  1.8 

P,o 

-0.8 

+  1.8 

P,, 

—  0.8 

+  1.8 

To  find  the  stress  in  G/&,  the  load  covers 
the  river  arm  and  central  span  to  the  right 
of  section  or -apex  points  h,  i,j,  Jc  and  Z,  are 
loaded  each  with  20  000  pounds,  except  that 

at  I,  which  is  one  half  or  2Q^QQQ.     Then  by 

2 

the  use  of  the  table  R,  is  found  to  be 


52 

20  000  (—  .4  —  .6  —  8,  --  8,)  +  10  000 
X  —  .8  =  -  -  60  000.  and  R8  is  20  000 
( +  1.4  +  1.6  +  1.8  4-  1.8 )  +  10  000 
X  + 1.8  =  +  150  000  then  V  =  —  60  000 
4-  150000  =  4-  90000  and  stress  in  Qh 
=  90  000  X  1-18  =  106  200  pounds,  the 
same  value  as  that  found  by  the  other 
method. 

This  method  has  its  greatest  advantage 
when  the  loads  are  unequal,  or  when  a 
uniform  live  load  with  excess  loads  is  used, 
as  will  be  shown  in  the  discussion  of  live 
load  in  railroad  bridges. 

ARTICLE  15.  —  SNOW   LOAD   AND   SNOW 
LOAD  STRESSES. 

In  addition  to  dead  and  live  loads,  high- 
way bridges  are  subjected  to  another  kind 
of  vertical  load,  at  times  in  certain  climates  j 
namely,  snow  load.  This  varies  according 
to  climate  from  0  to  20  pounds  per  square 
foot  of  floor  surface  (see  Roofs  &  Bridges, 
Merriman's  and  Jacoby,  Part  I,  Art.  41.) 
Snow  load  is  assumed  to  be  distributed 
uniformly  over  the  floor  surface,  and  con- 


53 

sequently  acts  on  the  members  of  the 
truss  in  the  same  manner  as  the  dead 
load. 

This  fact  makes  the  calculation  of  snow 
load  stresses  an  easy  matter,  if  the  dead 
load  stresses  are  known. 

Let  iv  be  the  dead  load  and  w1  the  snow 
load  per  linear  foot  per  truss  ;  S  the  stress 
in  a  member  due  to  dead  load,  and  Sl  the 
stress  in  the  same  member  due  to  snow 
load,  then 

™  =  S  andS'  =  S  *..  ...(15) 

' 


To  find  the  stresses  due  to  snow  load, 
multiply  the  dead  load  stresses  by  the 
ratio  between  dead  and  snow  load.  It  is  to 
be  borne  in  mind  that  the  dead  load  always 
acts,  while  the  snow  load  may  or  may  not 
act  ;  and  although  the  stresses  are  of  the 
same  nature,  they  are  kept  separate  in 
order  that  the  maximum  and  minimum 
stresses  due  to  dead,  live,  snow  and  wind 
loads  combined  may  be  determined,  as  is 
shown  in  the  table  of  maximum  and  mini- 
mum stresses  at  the  end  of  the  chapter. 

Assuming    the    snow    load    to    be    15 


54 


pounds  per  square  foot  of  floor  surface, 
the  snow  load  per  linear  foot  per  truss  for 
the  cantilever  of  Article  10  is  if  the  dis- 
tance of  trusses  apart  be  16  feet  and  there 
are  two  sidewalks  each  5  feet  wide 

4.  -i      e^                  15(16  +  5  +  5) 
outside  of  the  trusses,  — 2 T — 

2 

390 

-—  or  say  200  pounds  per  linear  foot  per 

truss. 

The  stress  in  A  &  due  to  snow  load  is  S' 

=  S  ~  =  16960  ^  ==  6780  pounds. 
w  500 


ARTICLE  16. — STRESSES  DUE  TO  WIND. 

To  counteract  the  effect  of  wind,  which, 
acting  horizontally,  tends  to  deflect  the 
truss  in  a  horizontal  plane,  just  as  the  ver- 
tical forces  tend  to  deflect  the  truss  in  a 
vertical  plane,  members  called  struts  are 
introduced,  extending  from  the  chord 
apex  point  of  one  truss  to  the  same  apex 
point  of  the  other  truss,  and  also  tension 
members  or  tie -rods  extending  from  the 
chord  apex  point  of  one  truss  to  the  next 


55 


apex  point  of  the  other  truss.  The  ar- 
rangement of  the  members  of  this  lateral 
system  is  like  the  members  of  a  Pratt 
truss,  as  shown  in  Fig.  xix. 

Upper  Lateral  System 
A'     B'    cf    p'    E'    F'    G'     H'    i'     j'    K'     i/ 

IXIXLXIXJXJXIXIXDxJXCKI 

ABCDEFGHi        J        KL 
ABC        DEFGHI         J        KL 


N 

7 

\/f 

X 

/ 

/N 

\ 

\ 

/\ 

& 

a 

d e *J~Q      h 

LXIXIXXXJXlXXIXXlXf 

Abode       / V      h      i      J       HI 
Lower  Lateral,  System 
Fig.  XIX 

The  wind  blowing  in  one  direction  stresses 
one  system,  and  blowing  in  the  opposite 
direction  stresses  the  other.  This  arrange- 
ment causes  the  diagonal  member  to  take 
tension  only. 

The  actual  surface  exposed  to  wind  in 
the  cantilever  bridge  is  an  unknown  quan- 
tity before  the  bridge  is  designed ;  there- 
fore, some  approximate  value  must  be 
assumed  in  order  that  the  stresses  due  to 


56 

wind  may  be  calculated.  A  closely  ap- 
proximate wind  load  is  found  for  simple 
trusses  (see  Merriman's  Roofs  and  Bridges 
Part  I,  Art.  42,)  by  assuming  the  mem- 
bers of  the  truss  to  be  each  one  foot  wide ; 
then  the  total  area  exposed  to  wind  is 
twice  as  many  square  feet  as  there  are 
linear  feet  in  the  skeleton  outline  of  the 
truss.  The  pressure  per  square  foot 
exerted  by  wind  may  be  taken  at  about 
30  pounds,  although  a  pressure  as  high  as 
40  pounds  per  square  foot  is  sometimes 
taken. 

Wind  load  on  the  truss  is  taken  as  act- 
ing uniformly  over  the  entire  length.  It 
is.  therefore  similar  to  the  dead  load,  ex- 
cept that  it  acts  horizontally  and  produces 
tension  in  the  leeward  chords  and  com- 
pression in  the  windward  chords.  The 
stresses  in  the  horizontal  system  effected 
by  wind  are  calculated  just  as  the  stresses 
would  be  for  a  Pratt  truss  system.  The 
distribution  of  shears  and  moments  due  to 
wind  on  truss  is  represented  by  the  dia- 
grams of  shears  and  moments  due  to  dead 
load,  shown  in  Figs.  II,  in,  IV  and  v. 


57 

The  wind  on  the  upper  chord  apex 
points  is  transmitted  by  the  upper  lateral 
system  of  truss  shown  in  Fig.  XIX,  directly 
to  the  abutments  and  piers.  That  on  the 
lower  chord  is  transmitted  to  the  piers  at 
one  end,  and  to  the  abutment  at  the  other, 
by  means  of  the  inclined  end  posts. 

The  wind-load  stresses  given  in  the 
table  have  been  calculated  as  follows : 
(The  results  are  necessarily  an  approxima- 
tion, since  the  true  area  exposed  to  wind 
is  not  known.)  The  skeleton  outline  of 
the  cantilever,  Fig.  XIX  (dimensions  of 
which  are  given  in  Art.  10),  is  about  1070 
feet.  Assuming  the  wind  pressure  per 
square  foot  at  30  pounds,  the  total  wind 
pressure  on  one  truss  is 

1070  x  30  =  32  120  pounds. 
Assuming  two-thirds  to  be  applied  to  the 
upper  chord,  since  it  carries  the  floor  sys- 
tem, and   one-third   to   the   lower   chord, 
gives 

2  X  32  120 


3X11 


=  1946, 


,  1  X  32  120      07Q  -, 

and  — ; stt97o pounds 

o  X  11 


58 


respectively  for  the  upper  and  lower  chord 
apex  wind  loads.  To  be  on  the  side  of 
safety,  and  giving  at  the  same  time  better 
values  for  computation,  these  may  be  in- 
creased to  2000  and  1000  pounds  respect- 
ively. 

To  find  the  stresses  in  the  upper  lateral 
system  (see  Fig.  xix),  proceed  as  follows  : 

R2  X  100  =  8  X  4000  x  *|°+10  000  X  180. 

A- 

Ra  =  28  800  +  18  000  =  46  800  pounds, 
and  R]X100^4X4000X50  —  10000X80 

-  3  X  4000  X  40. 
RT  =  —4800. 

Let  6'  —  angle  which  diagonals  in  upper 
lateral  system  make  with  the  vertical;  then 
secant  6',  =  1.6  and  tan  6'  —  1.25. 

Stress   in  A'  B  equals  4800  X  1.6  = 
+  7680  pounds. 
Stress  in  C  D 

4800  X  40  +  4000  X  20  _ 


for  wind  West,  and  33  000  for  wind  East. 

For  C'  D  the  wind  blows  West  and  the 

shear  is  —4800  —  4000  —  4000  =  12  800 


59 


pounds.  This  multiplied  by  the  secant  of 
the  angle  C'  D  D'  gives  4-  12  800  X  1.6 
:  +  C'D  =  +  20  480  pounds. 

When  the  wind  blows  in  the  opposite 
direction,  or  East,  the  member  C  D'  is 
stressed  an  equal  amount  -f  20  480  pounds. 

The  effect  of  the  wind  on  the  upper 
chord  is  to  turn  the  bridge  over,  as  in 


70200 


7O2OO 


Fig.  XX 


Fig.  XX,  which  is  a  cross-section  of  truss 
shown  in  Fig.  xix  at  F/.     The  total  wind 


force  acting  at  F  is  46  800  pounds,  and,  if 
the  two  trusses  are  rigidly  connected  by 
members  F/7  and  F7/  and  struts,  this 
force  of  46  800  pounds  tends  to  produce 
rotation  about  the  point/7,  and  is  held  in 
equilibrium  by  a  downward  force  R2  of 
70  200  pounds.  The  equation  of  moments 
about/7  is 

-  46  800  X  24  +  70  200  X  16  =  0. 
The  same  result  obtains  if  the  center  of 
moments  .  be    considered    as   midway  be- 
tween Ra  and  R72,  which  latter  act  as  a 
couple. 

If  the  dead-load  reaction  R2  is  less  than 
the  value  R2,  the  downward  force  neces- 
sary to  produce  equilibrium,  the  bridge 
will  overturn.  R2  due  to  dead  load  is 
117  000  ;  therefore  there  is  a  good  factor 
of  safety  against  overturning,  due  to  the 
assumed  wind  pressure  of  30.84  pounds 
per  square  foot,  since  in  order  to  overturn 
the  bridge  the  wind  would  have  to  exert  a 
pressure  per  square  foot  equal  to 

117000X16X30.84 


46  800  X  24 


=  51.4  Ibs. 


61 

The  members  F'/  and  F/',  known  as 
cross-bracing,  are  designed  to  take  tension 
only.  The  stress  in  F'/is  found  as  fol- 
lows: Take  the  center  of  moments  at/% 
and  state  the  equation  of  moments. 

-  46  800  X  24  +  F7/  Xp  =  0 
p  =  16  X  cos/F/'  =  16  X  .83  =  13.28. 


Therefore,  F'/=  46  S®°*  24^+8458Qlbs> 
13.^8 

In  the  table  of  final  maximum  and  mini- 
mum stresses,  the  stresses  due  to  overturn- 
ing effect  of  wind  on  truss  are  not  given, 
and  are  omitted,  because  their  effect  is  so 
small  as  not  to  materially  change  the  final 
results.  The  stresses  due  to  overturning 
effect  of  wind  on  truss  and  train  are  given 
in  the  table  of  final  maximum  and  minimum 
stresses  in  a  railroad  cantilever  bridge  and 
the  method  of  calculation  given  in  Art.  25. 

In  actual  practice  it  would  be  well  to 
compare  the  assumed  apex  wind  loads 
with  the  actual  wind  apex  loads  as  the 
result  of  multiplying  the  assumed  pressure 
in  pounds  per  square  foot  by  the  actual 
surface  exposed  in  the  designed  structure. 


62 


This  should  be  done  at  least  to  make  sure 
that  the  assumed  wind  apex  loads  are  on 
the  side  of  safety. 


Stresses  in  Lateral  Systems  due  to  Wind. 


Member. 

Wind 
East. 

Wind 
West. 

Member. 

Wind 
East. 

Wind 
West. 

A  A' 

—     8800 

—     3800 

H'l 

4-  22  400 

0 

BB' 

-     681  >0 

—     6800 

I'  J 

4-  16  UOO 

0 

CC' 

—  10  800 

—  10  800 

J'K 

+     9200 

0 

DD' 

—  14  800 

—  14  800 

K'L 

4      3200 

0 

EE' 

—  1H  800 

—  18  800 

A'  b 

0 

4     4870 

FF' 

—  44  800 

-  44  }*00 

Ab' 

4   '  4870 

0 

GG' 

—  20  OuO 

—  20  (  00 

b'  c 

0 

4-     6640 

HH' 

—  10  000 

—  16  COO 

be' 

+     6640 

0 

IF 

—  12  000 

—  12  OUO 

c'  d 

0 

4-10240 

J  J' 

—     501,0 

—     5000 

cd' 

+  10  240 

0 

XK' 

—  .    4000 

—     4000 

d'  e 

0 

4  13  440 

LL' 

0 

—     2000 

de' 

4  13  440 

0 

A'B 

0 

4     7680 

e'f 

0 

+  16  640 

B'C 

0 

4  14080 

ef 

4  16  640 

0 

<y  D 

0 

-+-  20  480 

i  J' 

0 

4-    8000 

D'E 

0 

+  26  81-0 

i'  J 

4-    8000 

0 

E'F 

0 

+  33  280 

hi' 

0 

4-  11  200 

FG' 

0 

4-  35  200 

h'  i 

4-  11  200 

0 

GH' 

0 

4  28  800 

g  h' 

0 

4  14  200 

HI' 

0 

4  22  400 

(1'h 

4-  14  200 

0 

IJ' 

0 

4  16  0'  0 

fff' 

0 

4  17  600 

JK' 

0 

4-     92(0 

fff 

4  17  600 

0 

KL' 

0 

4-     3200 

b  b' 

—     34(0 

—     0400 

AB' 

4     7680 

0 

CC' 

-     5400 

—     5400 

BC' 

4  U  080 

0 

dd> 

—     7400 

—     7400 

CD' 

-j-  20  480 

0 

ee'f 

—     9400 

—     9400 

D  E' 

4-  26  880 

0 

—  22  400 

—  22  400 

EF' 

4-  33  280 

0 

9  9' 

—  10  000 

—  10  000 

F'G 

4  35  200 

0 

h  h' 

—    80CO 

—    8000 

G'H 

-j-  28  800 

0 

i  i' 

—    6000 

—     6000 

63 


ARTICLE  17.  —  FALSE  MEMBERS  INTRO- 
DUCED FOR  PURPOSES  OF  ERECTION. 

The  principal  advantage  that  the  can- 
tilever bridge  possesses  over  other  forms 
of  bridges,  the  suspension  bridge  excepted, 
consists  in  its  economy  of  erection  under 
unfavorable  conditions.  Comparatively 
little  false  work  is  required.  The  bridge 
is  erected  by  beginning  at  the  pier,  and 
building  out  on  both  the  shore  and  river 
arms  until  the  abutment  is  reached  on  one 
side,  and  connection  made  at  the  middle 
of  the  central  span  on  the  other. 

In  order  to  make  connection  in  the  mid- 
dle it  is  necessary  that  the  central  span  or 
a  part  of  it  be  made,  temporarily,  a  con- 
tinuation of  the  river  arm ;  by  means  of  false 
members  introduced  merely  to  support  the 
arm  and  the  necessary  apparatus,  etc.,  used 
in  erection. 

It  is  readily  seen  that,  in  the  case  of  the 
cantilever  shown  in  Fig.  VII,  the  compres- 
sion member  extending  from  i  toj  &ndj 
to  Jc  with  a  vertical  member  J  j,  would 


64 


make  the  central  span,  or  as  much  of  it  as 
is  necessary,  a  part  of  the  river  arm  of  the 
cantilever. 

This  change  in  the  arrangement  of  the 
members  causes  a  change  in  the  nature 
and  magnitude  cf  stress  in  some  of  the 
members  of  the  truss. 

What  this  change  is  remains  to  be  deter- 
mined; so  that,  if  necessary,  provision  may 
be  made  in  the  cross-section  of  the  mem- 
bers effected  to  safely  erect  the  bridge. 

Fig.  XXI  represents  the  skeleton  diagram 
of  Fig.  VII  changed  by  the  false  members 
ij,j  k  and  3j  being  introduced  for  purposes 
of  erection. 


I        J       K       L 


The  dimensions  cf  truss  are  the  same  as 
that  of  Fig.  vn,  and  dead  apex  loads  the 
same,  10  000  pounds ;  but  the  live  load  will 
be  assumed  to  consist  of  a  single  concen- 
trated weight  to  represent  a  traveler  used 


65 


in  erection.  This  will  be  taken  at  40  000 
pounds.  Secant  of  angle  which  K  I  and 
J  fc,  etc.  make  with  vertical  is  1.38. 

The  position  farthest  out  on  the  arm 
that  the  traveler  is  likely  to  occupy  is  at 
K,  since  all  the  members  L  I,  K  I,  k  I  etc., 
are  erected  with  it  in  that  position  and 
connection  made.  This  position  gives 
greatest  moment,  and  consequently  great- 
est stress  in  all  chord  members,  to  the  left. 

The  stress  in  L I  is  —  10  000,  pounds  due 
to  dead  load.  K  I  =  10  000  X  1.38  =  + 

13  800  pounds.    *l=™™™ 

9520  pounds.     K  k  =  — 10  000  —  10  000 

-  40  000  =     -  60  000   pounds,   which   is 
greater  than  the  maximum  stress  due  to 
dead  snow  and  live  load  as  given  in  table. 
J  k  =    -  60  000  X  1.38  =  +  82  800  pounds. 
.  i       .  .        ,  50  000  X  20  +  10  000  X  40 

-21- 

-  66  660  pounds,    i  J  =  70  000  X  1.38  = 
—  96  600  pounds.     When  the  traveler  is 

brought  over  the  point  I  the  stress  in  I  i  is 

-  40  000  +  10  000  =  —  50  000  pounds. 
No  change   takes   place   in   any   other 


66 


members  throughout  the  truss,  at  least  to 
the  extent  of  changing  the  maximum  and 
minimum  stresses  due  to  dead  snow  and 
live  load.  The  following  table  shows  what 
members  are  stressed  during  erection 
greater  than  when  subjected  to  dead  snow 
and  live  load. 

In  addition  to  the  stress  due  to  wind  on 
the  truss  there  may  be  stress,  due  to  wind 
on  the  traveler,  which  amounts  to  consider- 
able, depending  of  course  upon  its  position 
and  the  amount  of  surface  exposed. 

Possible  Stresses  During  Erection. 


Member 

Dead  Load. 

Traveler. 

Wind. 

Maximum 

LZ 

—  10  000 

0 

0               —   10000 

Kl 

+  13  800 

0 

0            !     +    13800 

Kk 

—  20  000 

—  40  000 

0 

—   60000 

J  k 

-|-  27  600 

+  55  200 

0 

4    82800 

Jj 

f  Assumed. 
\+     2000 

0 

0 

+       20UO 

J  i 

—  41  400 

—  55  200 

0 

—   96  600 

li 

—  10  000 

—  40  000 

0 

—   50000 

KL 

0 

0 

±      6400 

±        6400 

JK 

+     9520 

0 

±  12  800 

+    22320 

I  J 

-j-  42  900 

+  76  200     i     ±  19  200 

-f-  138  300 

kl 

—     9520 

0 

±      3200 

—   12720 

jk 

—  28  600 

—  38  100 

±      6400 

—   73100 

ij 

—  28  600 

—  38  100 

0 

—   66660 

The  work  of  erection  cannot  be  safely 


67 

carried  on  at  times  when  the  wind  blows 
at  a  high  velocity,  at  which  time  the  travel- 
er should  be  run  back  to  a  point  of  safety. 
On  this  account  no  allowance  has  been 
made  for  stress  in  the  members  due  to 
wind  on  the  traveler  placed  in  a  position 
to  effect  the  members  given  in  the  table. 


ARTICLE  18.— FINAL  MAXIMUM  AND  MIN- 
IMUM STRESSES. 

A  table  of  stresses  due  to  dead  load,  live 
load,  snow  load,  and  wind  on  truss  is  given 
for  the  cantilever  bridge  shown  in  Fig's 
•vii  and  xix,  and  the  final  maximum  and 
minimum  stresses  given  in  the  last  two 
columns. 

The  overturning  effect  of  wind  on  the 
truss  has  not  been  considered.  It  amounts 
to  but  little  any  way  and  would  not  change 
the  final  results  much  in  this  case.  But 
in  a  through  bridge  it  should  be  consid- 
ered. Impact  has  been  omitted  because 
of  its  complication.  Initial  tension  would 
enter  into  the  final  results  of  some  of  the 


68 


members.  All  of  these  omitted  forces  are 
mentioned  merely  to  call  attention  to 
them,  so  that  the  student  may  investigate 
the  subject  in  works  in  which  they  are 
treated. 

Attention  is  called  to  the  final  results,  as 
showing  in  some  members  the  reversal  of 
stress  from  tension  to  compression  and 
visa  versa. 


69 


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5 


C5O^,t-^OOO        C;O 

§«5^33S2      °S  °°° 

I  +  I  +  I  +7  i       +1 


M  J 


72 

CHAPTER  III. 
KAILROAD  CANTILEVER  BRIDGES. 


ARTICLE  19. — LOADS  IN  RAILROAD  CANTI- 
LEVER BRIDGES. 

In  railroad  bridges  the  loads  causing 
stress  in  the  members  of  the  truss  are 
dead,  live  and  wind.  Snow  load  is  not 
considered,  because  it  falls  through  be- 
tween the  ties. 

The  calculation  of  dead  load  has  been 
fully  discussed  in  Art.  4  for  highway 
bridges,  and  the  same  remarks  apply  here. 

The  wind  loads  in  railroad  bridges  differ 
considerably  from  those  in  highway 
bridges.  In  addition  to  the  effect  of  wind 
blowing  on  the  truss  the  wind  blowing  on 
the  train  is  considered,  both  as  regards  its 
effect  in  stressing  the  lateral  bracing,  and 
in  overturning  the  bridge  and  causing  addi- 
tional stress  in  the  leeward  truss  members. 
The  wind  on  the  truss  is  considered  as  a 


73 

moving  load,  and  care  should  be  taken  to 
place  the  train  in  the  same  position  that 
it  occupied  when  live  load  stress  was  found, 
so  that  the  stresses  due  to  the  different 
causes  may  be  properly  combined. 

(For  live  load  in  railroad  bridges,  see 
Art.  22.) 


ARTICLE  20.- 


REACTION  DUE  TO  DEAD 
LOAD. 


The  stresses  due  to  dead  load  in  a  rail- 
road cantilever  bridge  are  calculated  in 
precisely  the  same  way  as  for  a  highway 
cantilever  of  the  same  arrangement  of 
members. 


The  railroad  cantilever  bridge  used  in 
the  following  analysis  of  stresses  will  be 
that  shown  in  Fig.  XXII. 


74 


It  differs  from  those  so  far  considered 
in  that  it  has  three  points  of  support,  RI, 
R2  and  R8,  and  the  cantilever  would 
therefore  form  a  system  in  which  the 
strains  are  ambiguous  if  the  web  system 
were  continuous  from  end  to  end.  If  the 
diagonals  in  the  panel  between  R2  and  R3 
are  omitted,  this  ambiguity  disappears,  in- 
asmuch as  the  strains  transmitted  by  the 
remaining  members  of  that  panel  are  those 
due  to  moments,  and  the  shear  in  the 
panel  is  zero. 

The  equations  for  reactions  are  found 
as  follows : 

Let  Pj  be  the  resultant  of  all  the  loads 
011  the  shore  arm,  P2  the  resultant  of  all 
the  loads  on  the  river  arm  and  central 
span,  and  let  l\  and  mi  be  their  respective 
distances  from  R2  and  R8.  The  funda- 
mental principle  that  the  algebraic  sum 
of  the  vertical  forces  shall  equal  zero, 
gives 

Bt  +  R2  +  B,  —  Pi  —  P2  =  0 (16) 

and  since  the  shear  in  the  panel  between 
B8  and  R8  is  zero, 

P2  —  B.  =  0,  opB,  =  Ps (17) 


75 

That  is  to  say,  R3  equals  the  load  on  the 
right  of  R8;  therefore 

B,+B,  =  P,  ...............  (18) 

The  equation  of  moments  of  the  external 
forces  with  reference  to  point  at  reaction, 
R2,  is 

R!  I  —  Pi  I,  +  P2  (a  +  mO  —  R3  a  =  0, 
but  R3  =  P2  and  the  equation  reduces  to 

Pl  ?i  —  P2^i 


I 

The  moment  of  forces  with  reference  to 
point  RI  as  origin,  gives 


=  0  ............  (20) 

Substituting  in  equation  (16)  the  values 
of  R3  and  RI  as  given  in  equations  (17) 
and  (19),  gives 


V 

These  equations  are  general  for  this 
class  of  cantilever,  and  may  be  used  for 
both  uniform  and  concentrated  load,  with 
slight  modifications. 


76 


ARTICLE    21. — STRESSES  DUE  TO  DEAD 
LOAD. 

Let  the  single  track  deck  railroad  bridge 
shown  in  Fig.  xxill  have  the  following  di- 
mensions :  Length  of  shore  arm  180  feet, 
river  arm  120  feet  and  central  span  120 
feet;  the  panel  length  on  the  upper  chord 
15  feet,  except  the  panel  between  R2  and 
R8,  which  is  10  feet;,  the  depth  of  truss 
30  feet  and  distance  apart  of  trusses  16 
feet. 

Let  the  dead  apex  load  on  the  upper 
chord  be  assumed  at  12  000  pounds,  and 
the  apex  load  on  the  lower  chord  4000 
pounds. 

The  total  dead  load  is  then  356  000 
pounds. 

P2  =  172  000  pounds,  and  Pl  =  184  000 
pounds. 

Taking  a  full  apex  load  at  A  and  a,  the 
reactions  are  found  to  be  as  follows  : 

From  (17),  R3  =  P2  =  172  000  pounds. 

Equation  (16)  gives  R,  +  R2  +  R3 
=  P!  +  P2  =  356  000  pounds. 


77 


From  (18),  Rx  +  R2  =  P1  =  184  000 
pounds  and  RI  is  found  from  (19)  to  be 
Ri  X  180  =  184  000  X  90  —  108  000  X  60 

-  16  000  X  45  —  48  000  X  120, 
or  R!  =  -f  20  000  pounds. 

R8   =  P!  —  RX  =  184  000  —  20  000  = 
+  164  000  pounds. 

With  the  reactions  known,  the  calcula- 
tion of  stress  in  the  members  of  truss  due 
to  dead  load  is  a  comparatively  simple 
matter.  The  members  are  arranged  after 
the  manner  of  the  Baltimore  truss  with 
chords  horizontal.  . 

The  angle  6  which  the  inclined  web 
members  make  with  the  vertical  is  45 
degrees,  the  secant  of  which  is  1.41. 

The  stress  in  each  sub-vertical  B  6,  D  d, 
F/,  etc.,  is  equal  to  the  apex  load  that 
comes  upon  them,  which  in  the  case  of 
dead  load  is  12  000  pounds. 

All  the  members  A  6,  C  d,  E/,  Ts,  Tu, 
etc.,  are  stressed  alike  and  equal  to 

12000  secS,  or 
6000  X  1.41  =  8460  pounds. 


78 


79 


To  calculate  the  stress  in  ac,  pass  a 
section  cutting  three  members  B  C,  C  b 
and  a  c,  then  take  the  center  of  moments 
at  C  and  equate  the  moment  of  the  forces 
on  the  left  of  the  section  to  zero, 

a  c  X  30  4-  (20  000  —  16  000)  30 

—  12000X15  =  0, 
and  a  c  =  —  2000  pounds. 
For  the  member  D  E  proceed  in  the  same 
manner,  taking  the  center  of  moments  at 
c,  the  equation  is 

D  E  X  30  +  4000  X  30  —  12  000  X  15 
+ 12  000  X  15  =  0  from 

which  D  E  =  —  4000  pounds. 
The  stress  in  G/is  found  by  multiply- 
ing the  shear  in  the  section  cutting  it,  by 
Sectf. 
Shear  =  20  000  —  3  X  4000  —  6  X 12  000 

=  64  000 

and  G/  =  64  000  X  1.41  =  90  240  pounds. 
The  stress  in  M  N  is  found  by  taking 
the  center  of  moments  at  n  and  express, 
ing  the  moment  of  the  forces  on  the  right 
of  the  section,  thus 

M  N  X  30  =  60  000  X  120  +  48  000  X  60 
+  48  000  X  60 


80 

or  M  N  =  +  432  000  pounds. 
From    the    fundamental    condition    of 
static  equilibrium,  namely,  that  the  sum 
of    the    horizontal     components    of     the 
stresses  in  any  section  musi?  equal  zero, 

M  N  —  mn  =  km  =  np. 

In  calculating  the  stresses  for  the  mem- 
bers in  the  river  arm  it  is  best  to  consider 
the  forces  on  the  right  of  the  section  •  for 
example,  to  find  the  stress  in  P  Q,  pass 
section  cutting  P  Q,  P  q  and  p  r.  Take 
center  of  moments  at  r,  and  the  moment 
of  the  forces  on  the  right  is  made  equal 
to  P  Q  times  its  lever  arm,  or 

P  Q  X  30  =  60  000  X  60+  40  000  X  30 

-  12  000  X  15, 
which  gives  P  Q  =  +  154  000  pounds. 

ARTICLE  22. — LIVE  LOAD. 

The  live  load  generally  taken  for  calcu- 
lation of  stresses  in  the  members  of 
bridges  in  America,  consists  of  two  of  the 
heaviest  locomotives  in  general  use,  fol- 


81 


lowed    by   a   train    load    of    about   3000 
pounds  per  linear  foot. 

The  exact  solution  of  stresses  due  to- 
such  a  live  load  for  simple  trusses  is  given 
in  standard  works  on  stresses  in  framed 
structures.  The  work  involved  in  calcu- 
lating the  stresses  due  to  the  true-wheel 
load  method  is  considerably  greater  than 
that  required  by  the  use  of  uniform  train 
load  with  excess  loads;  and,  since  the  lo- 
comotive load  specified  by  different  rail- 
road companies  varies  considerably  in 
different  parts  of  the  country,  there  arises, 
on  the  part  of  bridge  building  companies 
a  general  desire  for  some  conventional 
method  of  treating  the  train  load  which 
will  give  easy  and  short  computations 
without  giving  results  materially  different 
from  the  true  ones. 

Very  close  approximations  to  the  actual 
wheel  loads  have  been  found,  and  used 
quite  extensively  in  bridge  computation. 

The  one  given  and  used  by  Prof. 
A.  J.  Dubois,  in  his  u  Framed  Structures  " 
involves  the  use  of  two  concentrated  ex: 
cess  loads  placed  50  feet  apart,  either 


82 

ahead  of,  or  in  the  middle  of  a  uniform 
train  load,  as  desired,  for  max.  shear  and 
moments. 

Another  method,  and  one  quite  exten- 
sively used  on  account  of  its  simplicity 
and  satisfactory  agreement  with  the  wheel 
load  method,  was  proposed  by  Geo.  H. 
Pegram,  in  Transactions  Am.  Soc.  C.  E., 
for  1886.  This  method  makes  use  of  one 
excess  load,  which  may  occupy  any  posi- 
tion in  the  uniform  train  load,  and  which 
may  be  conceived  as  rolling  across  the 
span  on  top  of  the  uniform  train  load.' 

In  the  following  analysis  of  stresses  the 
live  load  is  taken  as  consisting  of  a  uni- 
form train  load  and  one  concentrated 
excess  load,  except  when  the  train  is  di- 
vided so  as  to  occupy  two  different  por- 
tions of  the  bridge,  when  an  excess  load 
may  be  taken  with  each  part.  This  kind 
of  loading  is  adopted  because  it  is  easier, 
and  renders  the  analysis  of  stresses  much 
simpler  and  more  easily  understood,  while 
at  the  same  time  omitting  none  of  the 
principles  involved  in  the  exact  wheel-load 
method. 


83 

ARTICLE  23. — LIVE  LOAD  STRESSES. 

Assuming  the  excess  load  acting  on  one 
truss  to  be  20  000  pounds,  and  the  uniform 
train  load  at  2000  pounds  per  foot  or  2000 
X  15  =  30  000  pounds  apex  load,  the  live 
load  stresses  for  the  cantilever  of  Fig. 
xxin  are  found,  as  follows  :  Since  the 
live  load  consists  of  uniform  .load  and 
one  excess  load,  the  proper  position  of 
these  loads  to  give  maximum  positive  and 
negative  shears  and  moments  will  be 
found  by  reference  to  the  rules  already 
established  in  Articles  11,  12,  and  13. 

The  maximum  Live-load  stress  in  B  &, 
D  dj  etc.,  will  occur  when  the  uniform 
train  panel  load  and  excess  load  come 
upon  them,  and  is  30  000+20  000  =  50  000 
pounds. 

Let  the  chord  stresses  be  considered. 
In  Arts.  11  and  13  it  is  shown  that  the 
chord  members  in  the  shore  arm  are  sub- 
ject to  positive  and  negative  bending 
moment,  according  to  the  position  of  the 
live  load.  Maximum  positive  moment, 
producing  compression  in  upper  and  ten- 


84 

sion  in  lower  chord,  occurs  when  the  uni- 
form live  load  covers  the  entire  shore  arm, 
with  the  excess  load  at  the  center  of 
moments. 

Maximum  negative  moment,  producing 
tension  in  upper  chord  and  compression  in 
lower  chord  occurs  when  the  river  arm  and 
central  span  is  loaded,  with  the  excess  load 
at  the  end  of  the  river  arm.  The  upper 
chord  is  always  tension  and  lower  chord 
always*  compression  in  river  arm,  and  is  a 
maximum  for  this  particular  kind  of  truss 
and  arrangement  of  web  members  when 
the  live  load  covers  central  span  and  that 
part  of  river  arm  to  the  right  of  origin  of 
moments  with  the  excess  load  placed  at 
the  end  of  the  river  arm. 

The  central  span,  being  a  simple  truss, 
requires  no  discussion  as  to  proper  loading^ 
since  that  is  supposed  to  be  understood. 

The  greatest  tensile  stress  in  ac  due  to 
live  load  will  be  produced  by  positive  mo- 
ment, or,  when  the  uniform  live  load  covers 
the  shore  arm  and  the  excess  load  at  the 
center  of  moments,  which  is  at  apex  point 
C.  The  reaction  R]  for  this  position  of 


85 


load,  if  half  a  uniform  panel  load  is  as- 

L  *    •    T}        30  000  x   13 
snmed  to  come  at  A,  is  ril  -         — - — 

—  15  000  +  -  x  20000  =  196666 pounds. 
6 

and  a  c  X  30  =  (196  666  -  -  15  000)  30 
-  30  000  x  15  =  +  166  666  pounds. 

This  result  is  true  if  a  b  is  allowed  to 
take  compression  which  it  is  not,  because 
the  counter  b  c  comes  into  action  thus  re- 
ducing a  c  to  o. 

For  F  G  the  center  of  moments  is  at  e 
and  the  excess  load  at  E.  The  reaction 

R1  is  180  000  +  -  X  20  000    =    193  334 

pounds. 

Then  —  FG  =  —  BP  = 
(193  334  —  la  OOP)  60  —  30  OOP  (45  +  30) 

30 
and  F  G •  =  -  -  281  670  pounds. 

In  order  to  find  the  stress  in  the  chord 
members  of  the  shore  arm  due  to  negative 
moment,  the  river  arm  and  central  span 
must  be  covered  with  the  uniform  train 
load,  with  the  excess  load  at  the  end  of 
the  river  arm. 


86 


The  reaction  RI  due  to  this  position  of 
the  load  is  equal  to 

(135  OOP  +20000)  120  +  210000   x  60 
~" 


pounds,  or  RI  =  —  173  333  pounds. 

To  find  maximum  compressive  stress 
in  c  e  pass  section  through  D  E,  E  d  and 
c  e,  and  take  the  center  of  moments  at  E, 

then  ce    =    173  333  X  60   =  -346670 

oU 

pounds. 

In  the  same  manner 

k  m  X  30  =  173  330  X  6  X  30 
and  Jem   =   mn   =   np   =  —  1040000 
pounds. 

Let  a  few  of  the  chord  stresses  in  the 
river  arm  be  next  considered.  Here  the 
upper  chord  is  always  in  tension  and 
lower  chord  in  compression. 

Attention  is  called  to  the  fact  that 
a  greater  stress  can  be  obtained  in 
the  upper  chord  members,  if  the  central 
span  and  river  arm  is  loaded  with  the 
uniform  load  on  the  right  of  the  center 
of  moments  than  if  the  load  extends  up 
to  the  section  with  the  excess  load  in 


87 


either  case  at  the  end  of  the  river  arm. 
This  is  proved  by  the  two  following  equa- 
tions representing  the  stress  in  P  Q  for 
the  two  conditions  of  loading  alluded  to  : 

Passing  section  through  P  Q,  P  q  and 
p  r  and  taking  center  of  moments  at  r,  the 
equations  of  moments  when  load  extends 
to  section  is  P  Q  = 
155  OOP  X  60  +  90  OOP  X  30  —  30  OOP  X 15 

30 
or  P  Q  =  +  385  000  pounds. 

When  the  load  is  placed  on  the  right 
of  the  origin  of  moments ;  that  is,  up  to 
and  including  apex  point  S,  the  equation 
is 
T>  n         155  000  X  60  +  90  000  X  30 

^  ~130~ 

400  000  pounds. 

This  proves  that  the  stress  in  P  Q  is 
15  000  pounds  greater  when  the  load  ex- 
tends only  as  far  as  the  middle  of  the 
panel  to  the  right  of  the  center  of  mo- 
ments than  when  it  is  brought  up  to  the 
section. 

For  the  stress  in  p  r  take  the  center  of 
moments  at  P,  then 


155  000  X  90  +  150  000  X  45 
*r-'  -sr 

or'j?  r  —  —  690  000  pounds. 

The  calculation  of  stresses  in  the  web 
members  of  the  river  arm  involves  the 
very  same  principles  of  loading  that  were 
used  in  the  calculation  of  web  stresses 
in  river  arm  of  highway  bridge.  Art.  14. 

Take,  for  example,  the  member  R  s.  Its 
stress  is  equal  to  the  shear  in  the  section 
multiplied  by  1.41. 

The  maximum  shear  will  take  place 
when  all  the  load  possible  is  put  on  the 
right  of  the  section,  or  when  the  central 
span  and  river  arm  right  of  section  is 
loaded  with  uniform  load,  and  with  excess 
load  at  some  point  between  section  and 
end  of  river  arm. 

The  maximum  shear  in  section  is  then 
245  000  pounds,  and 

Us  =  245  000  X  1.41  ==  +  345450  pounds. 
To  find  the  stresses  in  the  web  members 
of  the  shore  arm  is  the  most  troublesome 
part  of  the  whole  problem,  but  with  care 
in  placing  the  loads  in  the  proper  position 
to  produce  the  greatest  possible  positive 


89 

and  negative  shears,  the  stresses  become 
readily  known  when  Rj  is  known. 

Take,  for  example,  the  member  E  d. 
The  greatest  possible  tensile  stress  in  this 
member  will  occur  when  the  river  arm 
and  central  span  is  loaded  with  uniform 
load  and  with  excess  load  at  end  of  river 
arm  and  the  shore  arm  covered  left  of  sec- 
tion. 

The  reaction  R,  due  to  this  loading  is 
from  formula  (19),  Rz  = 

155  OOP  x  120+210  OOP  x  60—90  OOP  X 150 

180 
=  _  98  335. 

The  shear  in  section  is  then  —  98  335  — 
90  000  =  188  355  and 
E  d  —  188  335  x  1.41  = :  +  265  550  pounds. 

This  result  is  obtained  on  a  rather  re- 
diculous  supposition,  in  that  the  load  on 
left  of  the  the  section  on  shore  arm,  though 
isolated  from  the  other  load  on  river  arm, 
is  assumed  to  come  into  the  desired  position 
without  any  locomotive  or  excess  load  to 
place  it  there.  A  more  reasonable  suppo- 
sition would  be  to  place  an  excess  load  at 


90 


the  head  of  the  uniform  load  left  of  the 
section  on  shore  arm. 

Finding  RI  by  means  of  formula  (19) 
gives, 

—  Ri  X  180  =  155  000  X  120  +  210  000 
X  60  +  90000  X  150  —  20  000  X  135, 
and  R,  =  —  83  335  pounds. 

The  shear  in  section  is  —  83  335  — 
110  000  =  —  193  335  and  E  d  =  193  335 
X  1.41  =  +  272  600  pounds. 

The  stress  c  d  is  equal  to  the  stress  in 
E  d  minus  the  stress  in  E  d  caused  by  the 
apex  load  at  D  or  c  d  =  272  600  —  35  250 
=  +  237  350  pounds. 

Since  c  d  supports  C  c  the  stress  in  C  c 
must  equal  the  vertical  component  of  the 
stress  in  c  d ;  therefore  C  c  equals  237  350 
•*•  1.41. 

or  G  c  =  —  168  330  pounds. 

The  stress  in  N  n   =   R8    =   P2  = 
395  000  pounds,  and  M  m  equals  R2,  but 
from  formula  (18),  R2  =  P,  —  R,. 

The  greatest  value  for  R2  will  occur 
when  the  bridge  is  covered  with  live  load. 
That  on  the  river  arm  and  central  span 
being  in  the  position  occupied  for  maxi- 


91 


mum  negative  moment  in  shore  arm,  while 

the  shore  arm  is  covered  with  uniform  live 

load  with  excess  load  at  M.     This  gives 

P,==—  (12x30000+200004-15000)  =  - 

395  000  pounds,  and 

Rj  X  180  =  155  000  X  120  +  210  000  X 

60  --  (11  X  30000)  90  --  15000  X  180 

or  Rj  =  -|-  6666  pounds,  and 

R8  =  M  m  =  —  395  000  —  6666  =  401  666 

pounds. 

The  maximum,  negative  live  load  stress 
in  c  d  and  d  E  is  produced  when  the  shore 
arm  is  loaded  on  the  right  of  a  section 
cutting  D  E,  d  E  and  c  e  with  the  excess 
load  at  E.  RI  due  to  this  position  of  the 
load  is 


20  000  =  103  335  and  c  d  =  d  E  =  103  335 
X  1.41  =  —  145  700  pounds. 

Here  is  a  member  which  shows  itself 
to  be  subject  to  alternate  tension  and 
compression  for  different  positions  of  the 
live  load,  which  is  an  objectionable  con- 
dition, and  can  be  avoided  by  the  introduc- 
tion of  a  counter  member  d  e.  which  will 


92 

prevent  the  members  c  d  and  d  E  from 
taking  compression. 

The  actual  effective  compressive  stress 
that  can  occur  in  c  d,  is  the  algebraic  sum 
of  the  stresses  in  c  d  due  to  dead  live  and 
wind  on  train  loads,  which,  taken  from  the 
table  of  stress  are  +  42300  --  145  700 
and — 11  320  respectively,  the  sum  of  which 
is  —  114  720  pounds. 

This  is  very  nearly  the  value  of  the 
stress  in  the  counter  de.  A  counter  is 
needed,  therefore,  in  any  panel  in  which 
the  live  load  and  wind  overturning  load 
negative  shear  exceeds  numerically  the 
dead-load  positive  shear.  By  reference 
to  the  table  of  final  maximum  and  mini- 
mum stresses,  it  is  readily  seen  that  the 
only  panels  which  need  counter  bracing  are 
the  first  three  at  the  end  of  the  shore  arm 
or  be,  d  e  and  fg.  In  practice  another 
panel  might  be  counter  braced  for  the 
sake  of  security. 

ARTICLE  25. — WIND  LOAD  STRESSES. 
Wind  blowing  on   a  bridge  produces  a 


93 


double  effect.  First, — it  has  the  effect  of 
stressing  the  members  of  the  lateral-sys- 
tem, and  thereby  producing  compression 
in  the  windward  chords  and  tension  in  the 
leeward  chords.  Second, — it  has  the  effect 
of  overturning  the  bridge.  This  latter 
effect  produces  an  additional  vertical  load 
on  the  leeward  truss,  and  consequently 
greater  stress  in  the  members  of  it,  while 
at  the  same  time  decreasing  the  stress  in 
the  members  of  the  windward  truss.  The 
change  of  stress  in  web  members  of 
trusses  due  to  overturning  effect  of  wind 
is  caused,  however,  by  the  wind  on  train 
or  live  load  alone,  while  the  chord  mem- 
bers are  effected  by  both  wind  on  train 
and  wind  on  truss. 

Let  the  wind  apex  load  on  both  the 
upper  and  lower  chords  due  to  wind  on 
truss  be  2000  pounds,  except  the  end 
apex  load,  which  is  1000  pounds. 

The  stresses  in  the  lateral  system  and 
chord  members  are  now  found  by  apply- 
ing the  principles  given  in  the  case  of 
highway  bridge,  Art.  16.  The  reactions 


Np 

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95 


are  found  by  reference  to  Fig.  XXIY,  to  be 
as  follows : 

For  the  upper  lateral  system  from 
formula  (18),  R8=:+56000,  and  for 
lower  lateral  system  R3  =  +16  000 
pounds.  From  formula  (19),  Rj  for  upper 
and  lower  lateral  system/  equals  —  3333 
and  +  8000  pounds  respectively.  These 
results  are  obtained  on  the  supposition 
that  the  wind  apex  loads  on  the  central 
span  are  all  transmitted  by  the  lateral 
systems  of  the  central  span  to  the  end  of 
the  river  arm,  and  then  acts  through  the 
lateral  system  of  the  upper  chord.  This 
is  a  rather  more  reasonable  supposition 
than  that  in  the  case  of  wind  in  the  high- 
way bridge  of  Art.  16,  where  the  wind 
apex  loads  on  the  lower  chord  of  the  cen- 
tral span  were  assumed  to  be  transmitted 
to  the  end  of  the  river  arm,  and  then  into 
the  lower  chord  of  the  river  arm  by  means 
of  the  inclined  transverse  bracing  J  i1  and 
J'  i.  See  Fig.  xix. 

Rj  for  upper  system  is  found  from 
formula  (21)  to  be  +  53  333  pounds,  and 
for  lower  system  +18  000  pounds.  To 


96 

find  the  stress  in  any  web  member  of  the 
upper  lateral  system  due  to  wind  on  truss, 
multiply  the  shear  into  the  secant  of  the 
angle  which  the  member  makes  with  the 
vertical.  For  N  O'  shear  is  52  000  pounds, 

and   Sec<9  =   ?|  =  1.4,  and  N  O'  =  52  000 
16 

X  1.4  =  +  72  800  pounds. 
The  same  stress  takes  effect  in  N'  O  when 
wind  is  reversed.  Since  these  members 
are  duplicates  the  stress  is  given  for  only 
one  system.  Stress  in  P  P'  is  simply  the 
shear  or  P  P'  =  -46  000  pounds. 

The  overturning  effect  of  wind  on  the 
truss  is,  in  the  case  of  a  cantilever  bridge, 
a  doubtful  quantity,  and  very  difficult  of 
satisfactory  determination.  It  is  perfectly 
evident  in  the  problem  at  hand  that  the 
wind  blowing  on  the  shore  arm  affects  the 
chord  stresses  in  connection  with  the  lat- 
eral bracing,  and  that  this  effect  is  trans- 
mitted by  the  lateral  system  to  the  ends 
of  the  shore  arm,  where,  by  means  of  the 
cross-frame  it  is  transmitted  directly  to 
the  abutment  and  pier.  The  wind  on  the 
river  arm  and  central  span,  or  that  part  of 


97 


it  acting  on  the  lower  chord,  has,  however, 
the  effect  of  twisting  the  river  arm,  and 
thereby  causing  some  additional  stress  to 
the  chord  members  of  the  central  span,  as, 
well  as  additional  stress  in  both  chord  and 
web  members  of  the  river  arm.  This 
change  of  stress  must  take  place  either 
when  the  train  is  on  the  bridge  or  when 
the  bridge  is  unloaded.  In  the  first  case 
the  overturning  effect  of  wind  on  train 
would  have  the  opposite  effect  to  the  wind 
on  truss  5  or,  in  other  words,  would  counter- 
act the  overturning  effect  of  wind  on 
truss.  In  the  second  case,  the  wind  blow- 
ing at  a  time  when  no  train  is  on  the 
bridge,  the  overturning  effect  of  which  on 
truss  would  give  stresses  which  combined 
with  the  dead-load  stresses  would  give  re- 
sults very  much  less  than  the  possible 
maximum  stresses  caused  when  the  bridge 
is  loaded.  For  these  reasons  the  stresses 
in  the  members  of  truss  due  to  the  over- 
turning effect  of  wind  on  the  truss  will  be 
omitted. 

The  overturning  effect  of  wind  on  the 
train,  however,  gives  additional  stresses  in 


98 


members  of  the  truss,  which,  acting  at  the 
time  when  the  live  load  acts,  should  be 
taken  into  account  to  give  the  maxim/urn 
stresses.  Assume  the  train  to  consist  of 
box  cars  10  feet  high,  and  the  wind  pres- 
sure per  square  foot  30  pounds.  This 
gives  300  pounds  per  linear  foot,  or  300 
X  15  =  4500  pounds  per  panel.  Taking 
the  center  of  pressure  of  the  wind  at  9.5 
feet  above  the  center  of  the  upper  chord, 
the  overturning  moment  at  each  panel 
point  is  then  4500  X  9.5  =  42  750  pounds. 
This  causes  an  additional  vertical  weight 
to  act  at  each  apex  point  of  the  leeward 
girder,  equal  to  42  750  -4-  16  =  2675 
pounds,  and  relieves  the  windward  girder 
by  the  same  amount. 

This  apex  load  effects  the  chord  and 
web  members  of  the  truss  in  the  same 
manner  as  a  live  apex  load,  and  the  pro- 
cess of  finding  the  stress  is  consequently 
a  repetition  of  that  for  live  load. 

Since  the  final  maximum  and  minimum 
stresses  are  the  result  of  combining  those 
stresses  caused  by  the  different  possible 
loading,  it  is  necessary  that  care  should 


99 

be  taken  to  get  the  stress  in  the  members 
due  to  overturning  effect  of  wind,  when 
the  live  load  occupies  the  same  position 
on  the  bridge  that  it  occupied  when  the 
live-load  stresses  were  calculated. 

The  stress  in  the  members  of  the  cross 
frames  of  the  bridge  have  not  been  calcu- 
lated, since  the  method  of  calculating 
them  has  been  explained  at  the  end  of 
Article  16,  in  highway  bridges,  and  differs 
in  the  railroad  bridge  only  in  the  addi- 
tional surface  exposed  to  the  wind  by  the 
train,  or  4500  pounds  per  panel. 

This  force  is  to  be  considered  only 
when  the  bridge  is  a  deck  structure,  since 
the  wind  on  the  train  is  transmitted 
through  the  wheels  and  track  to  the 
chord  on  which  it  rests. 


100 


Section  of  Truss  and    Train    with    Forces    and 
Lever  Arms. 


Fig.  XXV 


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and        ' 


Ey  Pi  'LIVER  J,  LODGE.    Revised  Edition.    i2mot 

c^th,     illustrated.    ..............   .  ^i.co. 

Applied  Mechanics. 

A  Treatise  for  the  Use  of  Students  who  have  time  to 
Work  Experimental,  Numerical^  and  Graphical  Exercises, 
Illustrating  the  Subject    By  JOHN  PERRY,  M.  E.,  D.  Sc.,  , 
F.  R.  S.     8vo,  cloth,  ............  Price  $2.50.  ' 


